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Digitized by Google Digitized by Google Digitized by Google Digitized by Google ANALYTIC GEOMETRY Digitized by Google JlfeQra&'j/illBoQk G±7n& PUBLISHERS OF &OOKS FOJO Coal Age ▼ Electric Railway Journal Electrical W>rld * Engineering News-Record American Machinist v Ingenierfa Internacional Engineering' 8 Mining Journal * Power Chemical 6 Metallurgical Engineering Electrical Merchandising Digitized by Google ANALYTIC GEOMETRY WITH INTRODUCTORY CHAPTER ON THE CALCULUS BY CLAUDE IRWIN PALMER ASSOCIATE PROFESSOR OF MATHEMATICS, ARMOUR INSTITUTE OF TECHNOLOGY AND WILLIAM CHARLES KRATHWOHL ASSOCIATE PROFESSOR OF MATHEMATICS, ARMOUR INSTITUTE OF TECHNOLOGY First Edition McGRAW-HILL BOOK COMPANY, Inc. NEW YORK: 370 SEVENTH AVENUE LONDON: 6 6 8 BOUVERIE ST., E. C. 4 1921 Digitized by Google Copyright, 1921, by the McGraw-Hill Book Company, Inc. THK MAPI! PKBSS TOSKP1 Digitized by Google V / .!:•■;,' r-i PREFACE The object of this book is to present analytic geometry to the student in as natural and simple a manner as possible without losing mathematical rigor. The average student thinks visually instead of abstractly, and it is for the average student that this work has been written. It was prepared primarily to meet the requirements in mathematics for the second half of the first year at the Armour Institute of Tech- nology. To make it adaptable to courses in other institutions of learning certain topics not usually taught in an engineering school have been added. While it is useless to claim any great originality in treat- ment or in the selection of subject matter, the methods and illustrations have been thoroughly tested in the class room. It is believed that the topics are so presented as to bring the ideas within the grasp of students found in classes where mathematics is a required subject. No attempt has been made to be novel only; but the best ideas and treatment have been used, no matter how often they have appeared in other works on the subject. The following points are to be especially noted: (1) The great central idea is the passing from the geometric to the analytic and vice versa. This idea is held consistently throughout the book. (2) In the beginning a broad foundation is laid in the algebraic treatment of geometric ideas. Here the student should acquire the analytic method if he is to make a success of the course. (3) Transformation of coordinates is given early and used frequently throughout the book, not confined to a single chapter as is so frequently the case. The same may be said of polar coordinates. v M305O99 vi PREFACE (4) Fundamental concepts are dealt with in an informal as well as in a formal manner. The informal often fixes and clarifies the ideas where the formal does not. (5) Numerous illustrative examples are worked out in order that the student may get a clear idea of the methods to be used in the solution of problems. (6) The conic sections are treated from the starting point of the focus and directrix definition. (7) Because of its great importance in engineering practice the empirical equation is dealt with more completely than is usual. This treatment has been made as elementary as possible, but sufficiently comprehensive to enable one to solve the average problem in empirical equations. (8) The fundamental concepts of the calculus are presented in a very concrete manner, and a much greater use then is usual is made of the differential. The ideas are thus more readily visualized than is possible otherwise. The applications are mainly to tangents, normals, areas, and the discussion of equations. (9) The concluding chapter gives an adequate and careful treatment of solid geometry so necessary in the study of the calculus. (10) The exercises are numerous, carefully graded, and include many practical applications. (11) In the introductory chapter are found various short tables and formulas, and at the end are given four place tables of logarithms and trigonometric functions. The authors take this opportunity to express their indebted- ness to their colleagues, Professors D. F. Campbell, H. R. Phalen, and W. L. Miser, for their assistance in the preparation of the text. The Authors. Chicago, III., May, 1921. Digitized by Google CONTENTS CHAPTER I Introduction Abt. Pao« 1. Introductory remarks 1 2. Algebra and geometry united 1 3. Fundamental questions 1 4. Algebra • 2 5. Trigonometry 3 6. Useful tables 5 CHAPTER II Geometric Facts Expressed Analytically, and Conversely 7. General statement. 8 8. Points as numbers, and conversely 8 9. The line segment 9 10. Addition and subtraction of line segments 10 11. Line segment between two points 11 12. Geometric addition and subtraction of line segments 11 13. Determination of a point in a plane 12 14. Coordinate axes 12 15. Plotting a point 13 16. Oblique cartesian coordinates 15 17. Notation 15 18. Value of a line segment parallel to an axis 16 19 Distance between two points in rectangular coordinates ... 17 20. Internal and external division of a line segment 19 21. To find the coordinates of a point that divides a line segment in a given ratio 20 22. Formulas for finding coordinates of point that divides a line segment in a given ratio 22 23. The angle between two lines 24 24. Inclination and slope of a line 25 25. Analytic expression for slope of a line 25 26. Formula for finding the slope of a line through two points . . 25 vii Digitized by Google viii CONTENTS Abt. Pagb 27. The tangent of the angle that one line makes with another in terms of their slopes 26 28. Parallel and perpendicular lines 27 29. Location of points in a plane by polar codrdinates 29 30. Relations between rectangular and polar codrdinates 32 31. Changing from one system of axes to another 33 32. Translation of coordinate axes 33 33. Rotation of axes. Transformation to axes making an angle <p with the original 34 34. Area of a triangle in rectangular coordinates 36 35. Area of any polygon 38 36. Analytic methods applied to the proofs of geometric theorems 39 CHAPTER III Loci and Equations 37. General statement 44 38. Constants and variables 44 39. The locus 45 40. The locus of an equation 45 41. Plotting an equation 46 42. The imaginary number in analytic geometry 47 43. Geometric facts from the equation 48 44. Intercepts • 49 45. Symmetry, geometric properties 49 46. Symmetry, algebraic properties 50 47. Extent 51 84. Composite loci 53 49. Intersection of two curves 54 50. Equations of loci 55 51. Derivation of the equation of a locus 56 CHAPTER IV The Straight Line and the General Equation of the First Degree 52. Conditions determining a straight line 59 53. Point slope form of equation of the straight line 59 54. Lines parallel to the axes 60 55. Slope intercept form 61 56. Two point form 62 57. Intercept form 62 Digitized by Google CONTENTS ix Art. Pagb 58. Normal form 63 59. linear equations 65 60. Plotting linear equations . . . . • 65 61. Comparison of standard forms 66 62. Reduction of Ax + By + C - O to the normal form .... 66 63. Distance from a point to a line 68 64. The bisectors of an angle 70 65. Systems of straight lines 71 66. Applications of systems of straight lines to problems 72 67. Loci through the intersection of two loci 75 68. Plotting by factoring 77 69. Straight line in polar coordinates 78 70. Applications of the straight line . 79 CHAPTER V The Circle and Certain Forms of the Second Degree Equation 71. Introduction 86 72. Equation of circle in terms of center and radius 86 73. General equation of the circle 87 74. Special form of the general equation of the second degree . . 87 75. Equation of a circle satisfying three conditions 88 76. Systems of circles 92 77. Locus problems involving circles 94 78. Equation of a circle in polar coordinates 96 CHAPTER VI The Parabola and Certain Forms of the Second Degree Equation 79. General statement. 98 80. Conic sections * . . 98 81. Conies 99 82. The equation of the parabola 100 83. Shape of the parabola 101 84. Definitions 102 85. Parabola with axis on the t/-axis 102 86. Equation of parabola when axes are translated 103 87. Equations of forms y 2 + Dx + Ey + F = and x* + Dx + Ey +F = O 106 88. The quadratic function ax 2 + by + c 107 89. Equation simplified by translation of coordinate axes .... 107 Digitized by Google X CONTENTS Abt. Pages 90. Equation of a parabola when the coordinate axes are rotated 109 91. Equation of parabola in polar coordinates Ill 92. Construction of a parabola 112 93. Parabolic arch 113 94. The path of a projectile 114 CHAPTER VII The Ellipse and Certain Forms of the Second Degree Equation 95. The equation of the ellipse . 117 96. Shape of the ellipse 119 97. Definitions 120 98. Second focus and second directrix 120 99. Ellipse with major axis on the y-axis 121 100. Equation of ellipse when axes are translated 123 101. Equation of the form Ax 2 + Cy* + Dx + Ey + F - . . . 125 102. Equation of ellipse when axes are rotated 127 103. Equation of ellipse in polar coordinates 129 104. Construction of an ellipse 129 105. Uses of the ellipse 131 CHAPTER VIII The Hyperbola and Certain Forms of the Second Degree Equation 106. The equation of the hyperbola 134 107. Shape of the hyperbola 136 108. Definitions 137 109. Second focus and second directrix 137 110. Hyperbola with transverse axis on the y-axis 137 111. Asymptotes 139 112. Conjugate hyperbolas 141 113. Equilateral hyperbolas 142 114. Equation of hyperbola when axes are translated 143 115. Equation of the form Ax* + Cy 2 + Dx + Ey + F = . . . 144 116. Equation of hyperbola when axes are rotated 146 117. Equation of hyperbola in polar coordinates 147 118. Construction of an hyperbola. 148 119. Uses of the hyperbola 150 Digitized by Google CONTENTS Xi CHAPTER IX Other Loci and Equations Abt. Pao» 120. General statement 164 121. Summary for second degree equations 164 122. Suggestions for simplifying second degree equations 167 123. Parabolic type 158 124. Hyperbolic type 159 125. The cissoid of Diocles 160 126. Other algebraic equations 161 127. Exponential equations . 163 128. Applications 164 129. Logarithmic equations 165 130. The sine curve 167 131. Periodic functions 168 132. Period and amplitude of a function 169 133. Projection of a point having uniform circular motion. Simple harmonic motion 170 134. Other applications of periodic functions 172 135. Exponential and periodic functions combined 172 136. Discussion of the equation • 174 137. Loci of polar equations 175 138. Remarks on loci of polar equations 177 139. Spirals 178 140. Polar equation of a locus 178 vl41. Parametric equations 180 142. The cycloid. . 182 143. The hypocycloid 183 144. The epicycloid 185 145. The involute of a circle 186 CHAPTER X Empirical Loci and Equations 146. General statement 188 147. Empirical curves 188 148. Experimental data 190 149. General forms of equations . K 191 150. Straight line, y = mx + b 191 151. The method of least squares 193 Digitized by Google xii CONTENTS Abt. Paob 152. Parabolic type, y = ex*, n > 195 153. Hyperbolic type, y = cx n , n < 197 154. Exponential type, y - ab* or y = oe** 197 155. Probability curve 199 156. Logarithmic paper 200 157. Empirical formulas of the type y *= a + bx + ex* + dx 8 + gx n 203 CHAPTER XI Poles, Polars, and Diameters 158. Harmonic ratio 206 159. Poles and polars 206 160. Properties of poles and polars 209 161. Diameters of an ellipse 210 162. Conjugate diameters of an ellipse 212 163. Diameters and conjugate diameters of an hyperbola .... 213 164. Diameters and conjugate diameters of a parabola . . . 213 165. Diameters and conjugate diameters of the general conic . . . 214 CHAPTER XII Elements op Calculus 166. Introductory remarks 216 167. Functions, variables, increments 216 168. Illustrations and definitions 220 169. Elementary theorems of limits 221 170. Derivatives 222 171. Tangents and normals 223 172. Differentiation by rules 225 173. The derivative when J (x) is x 226 174. The derivative when f(x) is c 226 175. The derivative of the sum of functions 226 176. The derivative of the product of two functions 227 177. The derivative of the product of a constant and a function 227 178. The derivative of the quotient of two functions 228 179. The derivative of the power of a function 228 180. Summary of formulas for algebraic functions 230 181. Examples of differentiation 231 182. Differentiation of implicit functions 233 183. Discussion of uses of derivative 235 184. Properties of a curve and its function 235 Digitized by Google CONTENTS xiii Art. Pagb 185. Curves rising or falling, functions increasing or decreasing 236 186. Maximum and minimum 237 187. Concavity and point of inflection 239 188. Relations between increments 242 189. Differentials 243 190. Illustrations 243 191. The inverse of differentiation 246 192. Determination of the constant of integration 247 193. Methods of integrating 248 194. Trigonometric functions 250 195. Derivatives of sin u and cos u 250 196. Derivatives of other trigonometric functions 252 197. f sin udu and f cos udu 253 198. Derivative of log«i* 254 199. Derivative of log w ' 255 200. Derivative oLa* and e u 255 201. Derivative of -u* 256 202. Illustrative examples 257 203. f^,f e u du, and f a u du 258 CHAPTER XIII Solid Analytic Geometry 204. Introduction .... 261 205. Rectangular coordinates in space 261 206. Geometrical methods of finding the coordinates of a point in space 263 207. Distance between two points 263 208. Coordinates of a point dividing a line segment in the ratio fito r« 264 209. Orthogonal projections of line segments 266 210. Direction cosines of a line 267 >211. Polar coordinates of a point 269 212. Spherical coordinates 270 213. Angle between two lines 271 214. Locus in space 274 215. Equations in one variable. Planes parallel to the axes .... 274 216. Equations in two variables. Cylindrical surfaces 274 217. Spheres 276 218. Surfaces of revolution 276 219. Equations of curves in space 278 Digitized by Google xiv CONTENTS Abt. Paob 220. Sections of a surface by planes parallel to the coordinate planes. 279 221. Projections of curves on the codrdinate planes 280 222. Surfaces in space 282 223. General equation of second degree 284 224. Ellipsoid 284 226. The hyperboloid of one sheet 285 226. The hyperboloid of two sheets 287 227. Elliptic paraboloid 288 228. Hyperbolic paraboloid 289 229. Cone 290 230. Equation of a plane 292 231. General equation of a plane 292 232. Normal form of the equation of a plane 293 233. Reduction of the equation of a plane to the normal form. . . 293 234. Intercept form of the equation of a plane 294 235. The equation of a plane determined by three conditions . 295 236. Angle between two planes 295 237. Distance from a point to a plane 296 238. Two plane equation of a straight line 298 239. Projection form of the equation of a straight line 298 240. Point direction form of the equation of a straight line, symmetrical form 299 241. Two point form of the equation of a straight line 300 Summary of Formulas 303 Four Place Table op Logarithms 308 Table of Trigonometric Functions 310 Answers 315 Index 341 Digitized by Google ANALYTIC GEOMETRY CHAPTER I INTRODUCTION 1. Introductory remarks. — Although it is not always possi- ble for a student to appreciate at the outset the content of a subject, it is well, however, to consider the object of the study, and to understand as far as possible its fundamental aims. 2. Algebra and geometry united. — Analytic geometry, or algebraic geometry, is a subject that unites algebra and geom- etry in such a manner that each clarifies and helps the other. Lagrange says: "As long as algebra and geometry travelled separate paths their advance was slow and their applications limited. But when these two sciences joined company, they drew from each other fresh vitality and thenceforward marched on at a rapid pace towards perfection. It is to Descartes 1 that we owe the application of algebra to geometry — an appli- cation which has furnished the key to the greatest discoveries in all branches of mathematics." 3. Fundamental questions. — The fundamental questions of analytic geometry are three. First, given a figure defined geometrically, to determine its equation, or algebraic representation. 1 Rene* Descartes (1596-1650) was one of the most distinguished philos- ophers. It was in pure mathematics, however, that he achieved the greatest and most lasting results, especially by his invention of analytic geometry. In developing this branch he had in mind the elucidation of algebra by means of geometric intuition and concepts. He introduced the present plan of representing known and unknown quantities, gave standing to the present system of exponents, and set forth the well known Descartes' Rule of Signs. His invention of analytic geometry may be said to constitute the point of departure of modern mathematics. 1 Digitized by Google 2 ANALYTIC GEOMETRY [§4 Second, given numbers or equations, to determine the geo- metric figure corresponding to them. Third, to study the relations that exist between the geo- metric properties of a figure and the algebraic, or analytic, properties of the equation. To pursue the subject of analytic geometry successfully the student should be familiar with plane and solid geometry, and should know algebra through quadratic equations and plane trigonometry. While parts of analytic geometry can be applied at once to the solution of various interesting and practical problems, much of it is studied beeause it is used in more advanced subjects in mathematics. Some of the more frequently used facts of algebra and trig- onometry are given here for convenience of reference. 4. Algebra. — Quadratic equations. — The roots of the quad- ratic equation ax 2 + bx •+ c = are —b + Vb 2 - 4ac , m —b — y/b 2 — 4ac n = — • 7p , and r 2 = = 2a 2a n + f2 = 9 and rif2 = — a a These roots are real and equal if 6 2 — 4ac = 0, real and unequal if b 2 — 4ac>0, imaginary if 6 2 — 4ac<0. The expression b 2 — 4ac is called the discriminant of the quadratic equation. Logarithms. (iy log M N = log M + log N. (2) log (M -5- N) = log ilf - log JV. (3) log N n = n log AT. Digitized by Google §5] INTRODUCTION 3 (7) a 10 *" = N. (8) log ^ = -log AT. (10) log* clog. 6 = 1. " j (11) log, N = 2.302585 log i0 N. (9) log* N = j^-£ log. N. (12 ) i ogl0 N = 0.43429 log. N. The base e = 2.718281828459- • -. t = 3.141592653589- • •. 6. Trigonometry. — Formulas. (1) 2t radians = 360°, «■ radians = 180°. lSfl° (2) 1 radian = — = 57.29578° - - 57° 17' 44.$". IT (3) 1° = j|q = 0.0174533 - radians. (4) sin 2 + cos 2 = 1. (5) 1 + tan 2 = sec 2 0. (6) 1 + cot 2 = esc 2 0. 1 , „ 1 (7) sin = ^, and esc = . _ v ' esc 0' sin (8) cos = — £, and sec = sec 0' cos (9) tan = —r-z y and cot = z — 5 - v ' cot0 tan0 (10) tan = - = s - v ' COS0 C8C0 / 11N , - COS0 CSC0 (11) COt = - — ^ = ■=• ' sin sec (12) sin (a + B) = sin a cos B + cos a sin B. (13) cos (a + B) = cos a cos B — sin a sin 0. (14) sin (a — B) = sin a cos B — cos a sin 0. (15) eos (a — B) = cos a cos B + sin a sin 0. /i£!\ x / i o\ tan a + tan (16) tan (a + B) = = t r — ^* v ' 1 — tan a tan B fi*\ x / x>\ tana — tan £ (17) tan (cr - 0) = r - 1 — — -^=. — * 1 + tan a tan (18) sin 20 = 2 sin cos 0. (19) cos 20 = cos 2 - sin = 1 - 2 sin 2 = 2 cos 2 0-1. /oa\ x o/i 2 tan (20)ten2 ' = l-^taF? Digitized by VjOOQIC ANALYTIC GEOMETRY [(5 (21) sin \0 t — C08 (23) tan *--±>/nF (22) cos — cos 1 — cos »-±£ + cos 6 2 sin (24) (25) (26) (27) (28) (29) (30) (31) (32) (33) (34) (35) (36) (37) (38) cos B sin B 1 + cos B sin a + sin = 2 sin J(a + 0) cos J(a — 0). sin a — sin = 2 cos J(a + 0) sin J(a — 0). cos a + cos = 2 cos J(a + 0) cos J(a — 0). cos a — cos = —2 sin J(a + 0) sin \{a — 0). sin a cos = J sin (a + 0) + i sin (a — 0). cos a sin = \ sin (a + 0) — J sin (a — 0). cos a cos = J cos (a + 0) + J cos (a .— 0). sin a sin = —J cos (a + 0) + J cos (a — 0). (Sine Law.) (J* — 0) = cos 0. (§* — *)" sin •• (£* - 0) m cot 0. ( Jt - 0) = tan 0. (§* + •) — cos 0. (§* + •) = —si 11 •• (Jr + 0) = -cot0. (J* + •) = -tan0. ( * — 0) = sin0. ( ir — 0) = —cos 0. ( it — 0) « — tan0. ( ir — 0) = — cot0. ( ir + 0) = — sin0. ( *■ + 0) = — COS0. ( ir,+ 0) = tan0. ( ir + 0) = cot 0. (■§tt — 0) = — cos0. (fir - 0) = -sin0. (|tt — 0) = cot 0. ($* - 0) » tan 0. (Cosine Law.) Digitized by Google §6) INTRODUCTION (39) sin (frc + 0) « -cos 0. cos (-fir + 0) = sin 0. tan (fir + 0) = — cot0. cot (fir + 0) = — tan0. (40) sin (2ir - 0) = -sin 0. cos (2* — 0) = cos 0. tan(2ir — 0) = — tan0. cot (2* - 0) = -cot0. (41) sin ( — 0) = — sin 0. cos ( — 0) = cos 0. tan ( — 0) = — tan 0. cot ( — 0) = — cot0. 6. Useful tables. Values of e* FROM x -0 TO X -4.9 X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00 1.11 1.22 1.35 1.49 1.65 1.82 2.01 2.23 2.46 1 2.12 3.00 3.32 3.67 4.06 4.48 4.95 5.47 6.05 6.69 2 7.39 8.17 9.03 9.97 11.0 12.2 13.5 14.9 16.4 18.2 3 20.1 22.2 24.5 27.1 30.0 33.1 36.6 40.4 44.7 49.4 4 54.6 60.3 66.7 73.7 81.5 90.0 99.5 109.9 121.5 134.3 Values of t r* FROM X = TO X = 4.9 X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00 0.90 0.82 0.74 0.67 0.61 0.55 0.50 0.45 0.41 1 0.37 0.33 0.30 0.27 0.25 0.22 0.20 0.18 0.17 0.15 2 0.14 0.12 0.11 0.10 0.09 0.08 0.07 0.07 0.06 0.06 3 0.05 0.05 0.04 0.04 0.03 0.03 0.03 0.02 0.02 0.02 4 0.02 0.02 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 Digitized by Google ANALYTIC GEOMETRY [§6 ?H ^ 1-4 QQ 1 04 1 OQ 1 04 o OQ Vcsc* - csc > ?H ^ 1 «> Oi 1 « 1 ^ OQ 04 ^ QQ > > > > 04 <u <£> +» ■+3 8 * + o 1 O o cot ot o o ?H i-4 « o > > 1-4 > > 1—1 > C3 + i-H > «> «A 04 A »A oj 0* +a ■+J + + l-H i—l > > «A «A <£> W 04 8 *> o g oo O QQ 5S s 2 ~* 1 § 1 o o 1 o ,_4 1—1 1-4 > V > > «a <£> «i «A 04 a a Q 'QQ *> QQ QQ ON QQ 1 ti ^ t-4 , *"* 1 »5 • as 1 1 A 1 l-H »H »H QQ »H > > > > C3 03 Digitized by Google §6] ^■to^INTRODUCTION 7 Table of Frequently Used Trigonometric Functions 6° in radians sin0 cos tan cot sec CSC $ 0° 1 00 1 00 30° 5 1 2 V3 2 . V3 3 V5 2\/5 3 2 45° 4 V2 2 2 1 1 V5 V5 60° 5 V3 2 1 2 V3 3 2 2V3 3 90° 2 1 00 00 I 120° 2t 3 V3 2 1 2 -V5 V3 3 -2 2VS 3 135° 3t 4 V2 2 V2 2 -l -1 -V5 V5 150° 5t 1 V3 V3 -V3 2VS 2 6 2 2 3 3 180° T -1 00 -1 00 210° 7x 6 1 2 V3 2 V3 3 vs 2\/3 3 -2 225° 5t 4 V2 2 V2 2 1 1 -V5 -V5 240° 4x 3 V3 2 1 2 V3 3 -2 2\/3 3 270° 3t 2 -1 00 00 -1 300° 5t 3 V3 2 1 2 -V3 V3 3 2 2\/3 3 315° 7x ^4~ V2 2 V2 2 -1 -1 VS -V2 330° llx 6 1 2 V3 2 V3 3 -V3 2a/3 3 -2 360° 2x 1 00 1 00 Digitized by Google CHAPTER II GEOMETRIC FACTS EXPRESSED ANALYTICALLY, AND CONVERSELY 7. General statement. — Geometry deals with points, lines, and figures composed of points and lines. Algebra deals with numbers and algebraic statements composed of numbers, such as the equation. In order to study geometric relations by means of algebra, and conversely, it is necessary to be able to represent points, lines, and geometric figures by means of numbers and equa- tions, and conversely. That is, it is necessary to be able to translate from the language of geometry to that of algebra, and conversely. 8. Points as numbers, and conversely. — If a point moves from A to £ in a straight line, the point is said to generate the line segment AB, that is, the line segment AB. is the locus of the point. If the point moves from B to A it generates the line segment BA. It is con- — ■ — *? venient to consider AB and BA as Pjq x separate line segments having oppo- site directions. The arrow is often used to denote the positive direction. Such line segments as AB and BA are called directed line segments. The point from which the moving point starts is called the initial point, and the point where it stops is called the terminal point. It is to be noted that a line segment is read by naming the initial point first. Let X'X be a straight line of indefinite length, and choose: first, a unit of length; second, a direction of motion, which we shall call positive if toward the right and negative 8 Digitized by Google §9] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 9 if toward the left; third, a point called the origin from which to start. Then any poiqt P can be determined by a real number — inn tegral, fractional, or irrational — which shows the number of units the point has moved from the origin. The number is positive or negative according as the motion is in the positive or negative direction. The origin is desig- nated by 0. Conversely, any real number corresponds to a point which is distant that number of units in the proper direction from the origin. Thus, in Fig. 2, +6 designates the point Pi and —2 the point Pi; while R corresponds to — 3^, and Q to V3. The line X f X is a directed straight line if it is thought of as generated by a point moving in the direction from X 9 to X or from X to X'. .Unit. _/ , , J* Pt O Q P P, *— j 1 J—M 1 1 1 1 H 1 ^— H H*X -6-6-4-3-2-1 1^2 3 4 5 « Fig. 2 9. The line segment. — The magnitude of a line segment is determined by the number of units in its length, that is, by the number of units a point moves in generating it. The value of a line segment is determined by its length and direction, and is defined to be the number which would represent the terminal point of the segment if the initial point were taken as origin. It follows from this definition that the value of a line seg- ment read in one direction is the negative of the value if read in the opposite direction. Thus, AB - -BA, or AB + BA - 0. By the numerical value of a line segment is meant the num- ber of units of length in it without reference to its direction. Two line segments are equal if they have the same direction and the same length, that is, the same value. Digitized by Google 10 ANALYTIC GEOMETRY [§10 In Pig. 3, AB - +2, CD - +2, ZW - +6, EC « -4, FA - -8„ AB — CD, and AF — CAT. AC and FD are equal in numerical value. A BC i> # F N , Unit t ♦— 1 * h-H* — » H » 1 1 H- *- »— I Fig. 3. EXERCISES 1. Draw a line segment 5 in. long and take the origin at the center. Choose as a unit of measure a line i in. long. What numbers designate the ends of the line? Locate the points corresponding to the numbers 9, 7i -4, -3i V2, - aA-t. 2. Draw a line segment 20 units in length, with the origin, 0, at the center. Locate the following points: A corresponding to 3, B corre- sponding to 8, C corresponding to — 4, D corresponding to — 10, E cor- responding to 10. Give the values of the following line segments: AB, DA, CE, BC, EA, AC. 3. In exercise 2, how are the numbers designating the points affected if the origin is moved two units to the right? How are the values of the line segments affected? L A C B M iy w % 1 1 1 1 1 H>- Fig. 4. 10. Addition and subtraction of line segments. — In Fig. 4, if A, B, C, • • • M, N are any arrangement of points on a ^ p P straight line, then •"£ ^ f AB + BC+ • - - +MN + NA-0. Pl Pt For the moving point generates in succes- * ' *" sion the line segments AB, BC, • • • MN, IG * NA, starting at A and returning to A. It therefore generates as much in the negative direction as in the positive. Hence the sum is zero. A case of frequent occurrence is that of three points 0, Pi, and P2 on a straight line, Fig. 5. If is taken as origin, then [1] (1) OP 2 = OPi + PxP2, (2) PJ> 2 - OP 2 - OPi. Digitized by Google §11] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 11 Proof. OPi + P X P 2 + P2O = 0. Why? Adding OP 2 = OP 2 , gives OP x + P1P2 = OP*. Adding OP 2 + PiO = OP* + PiO, gives PiP* =[OP 2 + P x O. .'. PiP 2 = OP 2 - OPu 11. Line segment between two points.— To find the value of the line segment between two points on a straight line, when the numbers determining these two points with reference to an origin on the same line, are known. In Fig. 5, is the origin and x x and x 2 are the numbers determining the points Pi and P 2 respectively. It is required to find the value of the line segment P1P2, that is, the magni- tude and direction of P1P2. P1P2 = OP 2 - OPi. By [1]. But OPi = xi, OP 2 = x 2 . [2] .*. P1P2 = x 2 - xi. This states that the value of the line segment between two points on a straight line is equal to the number determining the terminal point of the line segment minus the number determining its initial point, when a point on the straight line is taken as the origin. p 6 p 6 O Pi Pi Pz A 1 i 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ' 1 1 ' p -11-10-9 -8-7-6-5-4-3-2-1 1 2 » 4 5 7 8 9 10 U 12 13 FlQ. 6. Thus, in Fig. 6, PiP% = OP» - OP x - 5 - 3 = 2. PJPt = OPi - OP, = 5 - 8 - -3. PsPe - OP, - OP, = -4 - (-8) - +4. PJ>t - OPt - OP, - 5 - (-4) - +9. PzP, - OP, - OP, - -8 - 8 - -16. 12. Geometric addition and subtraction of line segments. From the preceding article it readily follows that two line seg- ments having the same or opposite directions can be added by placing the initial point of the second upon the terminal point of the first. The sum of the line segments is the line segment Digitized by Google 12 ANALYTIC GEOMETRY [§13 having, as initial point, the initial point of the first and } as terminal point, the terminal point of the second. A line segment is subtracted from another by reversing its direction and adding. Thus, in Fig. 6, OP x + PiP, - OP* - 8. Pft + PJ>> - PJ>> « -11. PJ>, - PtP* - PJ> % + PJ> t - PtPi - 9. PJ>< - P.P4 - PJ>* + P4Pt - PiPt - -9. EXERCISES 1. On a line with origin at 0, locate the following points: A determined by 2, B by 3, C by 8, Dby -5, E by -8. By the method of article 10, find the value of the line segments AB, BC, BD, AE, DE, EB, CE, CD. 2. On a line with origin at 0, locate the points Pi, P s , P s , Pa, determined by the numbers x x , x ty x 9 , x* respectively. (1) Give the values of the line segments PiP», PiP*, PiPj, P4PL (2) Give the line segments that have the following values: x* — x h X\ — Xt, x% — x t . Do the relative positions of the points make any difference in the answers? 13. Determination of a point in a plane. — It was shown in article 8 that the position of a point on a straight line can be determined by one number, which shows the direction and the distance that the point is from a fixed point on the straight line. Various methods may be given for locating a point in a plane. For the purposes of analytic geometry, two of these will be chosen. They correspond to the two methods ordinarily used in locating a point on the surface of the earth. First, a house in a city is located by giving its street and number. That is, by stating its distance and direction from each of two intersecting streets. Second, a city may be located by giving its distance and direction from another city. In analytic geometry, the two corresponding methods of locating a point in a plane are (1) the method by cartesian codrdinates, and (2) the method by polar coordinates. CARTESIAN COORDINATES 14. Coordinate axes. — (1) The lines of reference X'X'and Y'Y, Fig. 7, intersecting in the point 0, are chosen. These Digitized by Google §15] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 13 lines are considered perpendicular to each other in this article, and will always be so taken unless otherwise stated. The line X'X is called the axis of abscissas or the x-axis. The line Y f Y is called the axis of ordinates or the y-axis. To- gether they are called the coordinate axes. When the coordinate axes are perpendicular to each other they form a rectangular system. The coordinate axes divide the plane into four quadrants, numbered I, II, III, and IV as in trigonometry. (2) A line segment of convenient length is chosen for a unit of measure. This may be of any length whatever. (3) The direction is chosen as positive when towards the right parallel to the x-axis, or upwards parallel to the y-axis. p f Hence the negative direction is towards the left, or down- wards. 15. Plotting a point.— A x'-+ point Pi in the plane is de- *L M % O *i Nt *** *P* Fio. 7. termined by the line segments NiPi and Af 1P1, Fig. 7, drawn parallel to X'X and Y' Y re- spectively, for the values of these line segments tell how far and in what direction Pi is from the lines of reference. Here the line segment NiPi — +5, and M\P\ = +4. The point P% is determined by the line segments NtPi — — 3, and M,P S - +6. The point P 9 is determined by the line segments NtP 9 — — 6, and MsP, - -4. It is evident that any point in the plane is determined by one pair of numbers, and only one; and, conversely, every pair of real numbers determines one point in the plane, and only one. The two numbers that determine a point in a plane are Digitized by Google 14 ANALYTIC GEOMETRY 1515 palled the coordinates of the point. The number which is the value of the line segment parallel to the z-axis is called the abscissa of the point, and is usually represented by x. The number which is the value of the line segment parallel to the 2/-axis is called the ordinate of the point, and is usually repre- sented by y. The coordinates are written, for brevity, within parentheses and separated by a comma, the abscissa always being first, as (x, y). The letter designating the point is often written just before the parentheses. Thus, the points in Fig. 7 are written: Pi(5, 4), P*( -3, 6), P»(-6, -4), and p4(8, —3). The points Mi, Mi, Ni, N t , and are respectively the points (5, 0), (-3, 0), (0, 4), (0, -4), and (0, 0). It is evident that, in the first quadrant, both coordinates are positive; in the second quadrant, the abscissa is negative and the ordinate positive; in the third quadrant, both coordinates are negative; and, in the fourth quadrant, the abscissa is positive and the ordinate negative. When a point is located in a plane by means of its coordi- nates it is said to be plotted. The locating of points is v greatly facilitated by using paper that is ruled into small squares. Such paper is called coordinate paper. 1 4 t 1 JJH 4> 1 D 1< 5,3;) o p ? ( - t M) p,< B? h) \ ' 1 1 Example. — Plot the points Pi(5,3), P»(5, -3), P,(-2, -4), andP 4 (-4,4). The point Pi(5, 3) is plotted by counting off from along X'X a num- *ber of divisions equal to the abscissa 5, and then from the point so de- termined, a number of divisions on a line parallel to the ^-axis, equal to the ordinate 3. The points (5,-3), (-2,-4), and (-4, 4) are located in a similar manner. Y Fig. 8. Digitized by Google ++~x §16] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 15 16. Oblique cartesian coordinates. — In determining a point in a plane, it is not necessary that the coordinate axes shall be perpendicular to each other, but they may form an angle a). Such a set of axes is called an oblique cartesian system. In Fig. 9, the abscissa of Pi is NiPi - 3, and its ordinate is M iPi - 5. The coordinates of P t are N*Pt = — 4, and MiPj = 3. 17. Notation. — To secure clearness of statement, sub- scripts will be used with the letters designating points, and they will agree with the sub- scripts used with the coordi- nates of the points. Thus, the point Pi has coordi- nates (x h j/i), the point P s has coordinates (xt, y»), and so on. Points designated in this manner will, in general, be fixed points, while a point that may vary in position will be designated by a letter, as P, without a subscript and have coordinates (rr, y). EXERCISES 1. Draw a pair of axes and plot the following points:, (2, 3), (7, 9), (-2, 4), (-7, -2), (4, -3), (-2, -8), (0, 0), (0, 5), (-6, 0). 2. Draw the triangle whose vertices are (0, 2), (—2, —3), and (3, —2). 3. Draw the quadrilateral whose vertices are (3, 0), (0, 2), (— 6> 2), and(0, -2). 4* If the ordinate of a point is 0, where is the point? Where if its abscissa is 0? Using x for the abscissa and y for the ordinate, express each as an equation. 6. What is the locus of all points that have abscissas equal to 5? Of all points having ordinates equal to 10? Use x for the abscissas and y for the ordinates and write these statements as equations. 6. The abscissas of two points are each a. How is the line joining them situated with reference to the y-axis? The ordinates of two points are each —6. Sow is the line joining them situated with reference to the x-axis? Write each of these lines as an equation. 7. Two points are placed so that the abscissa of each is equal to Digitized by Google 16 ANALYTIC GEOMETRY [§18 its ordinate: How is the line joining the points situated with reference to the coordinate axes? In what two quadrants can the points lie? Write the equation. 8. Two points are placed so that the abscissa of each is equal to the negative of the ordinate. How is the line connecting them situated with reference to the coordinate axes? In what two quadrants can the points lie? Write the equation. 9. Draw a rectangle whose vertices are (—4, 2), (—4, —5), (7, —5), and (7, 2). Find the length of its sides by differences of abscissas or ordinates. 10. The vertex of a square is at the origin, and a diagonal lies on the positive part of the x-axis. # Find the coordinates of the other vertices if a side is 10. 11. What is the locus of a point which moves so that the ratio of its ordinate to its abscissa is always 1? So that this ratio is always — 1? Always 2? Write the equations. 12. An equilateral triangle of side a has a vertex at the origin and one side on the x-axis at the right of the origin. Find the coordinates of its vertices. 13. A regular hexagon of side 8 is placed so that its center is at the origin and one diagonal is along the x-axis. Find the coordinates of its vertices. 18. Value of line segment parallel to an axis. — If the segment of a line is parallel to one of the coordinate axes, it has a definite direction as well as a length, that is, it has a value. If P\{x\> yi) and P%(x 2 , y 2 ) are any two points on a line parallel to the x-axis, then [2i] P1P2 = x 2 - Xi. This follows directly from article 11, for if P1P2 intersects the y-axis in JYi, P1P2 = NiP 2 — N1P1 = x 2 — x\. Likewise, if PiOd, y{) and P 2 (x 2) 2/2) are any two points on a line parallel to the y-axis, then [2 2 ] P1P2 = y 2 - yi. The student should locate points in various positions and satisfy himself that [2J and [2 2 ] are true. Figure 10 shows several positions of Pi and P2 These facts may be stated as follows: Digitized by Google p* W k Pi Ni Pi r* / — *x §19] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 17 (1) The value of a line segment parallel to the x-axis equals the abscissa of its terminal point minus the abscissa of its initial point. (2) The value of a line segment parallel to the y-axis equals the ordinate of its terminal point minus the ordinate of its initial point. 19. Distance between two points in rectangular coordi- nates. — (1) The distance between two points is the numerical value of the line segment connecting these points, that is, it is the length of the line segment connecting the two points. It follows that the distance between two points, having abscissas X\ and x% on a line parallel to the 3-axis, is either x* — x\ or x\ — X2 f the differ- ence being taken positive when its^l numerical value can be determined. Likewise, when the two points p, p l are on a line parallel to the y-axis the distance between them is y 2 — FlG> 10 yioryi - y 2 . (2) Ordinarily a line segment that is not parallel to one of thfe coordinate axes does not have a direction assigned to it. We do not then speak of its value. The length of such a line segment is the distance between its end points. The distance between two points Pi(xi, y{) and P2O&2, yt) is given by the formula [3] d - V(x!-x 2 ) 2 + (yi-y2) 2 . Proof. — Let Pi(x h yi) and P 2 (x 2j 2/2) be the two points. Through Pi and P2 draw lines parallel, respectively, to the x-axis and y-axis to intersect in Q. Then P1QP2 is a right triangle, and d = PJ> 2 = Vp&* + QP a *. But PiQ = x 2 - x h and QP 2 = 2/2 - 2/1. By [2i] and [2 2 ]. Hence d = V(x 2 - *i) 2 + (2/2 - 2/1) 2 . ATi Digitized by Google 18 ANALYTIC GEOMETRY [510 Since (x 2 - Xi) 2 = (xi - x 2 ) 2 and (y 2 - j/i) 2 = (y x - j/ 2 )*, d = V(xi - x 2 ) 2 + (2/1 - y*)*. It should be noted that the line through Pi could as well have been drawn parallel to the Jt(*!.v») y-axis, and the line through P* parallel to the rr-axis. It is to be noted that the above proof is general and is made with- out reference to a figure. The student, however, should draw several figures locating the points in different positions and satisfy [3]. Figure 11 shows one position of ^ Pi(*i.*i) +X »i-*i Q(*«,lfi) Fig. 11. himself of the truth of the points. EXERCISES 1. Find the distance between each of the following pairs of points: (1) (3, 4), (-6, -8). (3) (-1, 0),(12, -2). (2) (-10, 4), (3, -9). (4) (6, 7), (-5, -5). *i(«**i> 2. In Fig. 12, express each of the fol- lowing line segments as the difference between two abscissas: M1M4, M*M* y 3. Express each of the following line segments as the difference between two ordinates: NiN h N*N h NJfi, NtN*. 4. Derive the formula for the distance between Pi(x h y x ) and Pi(xt t y»), (1) when both Pi and Pi are in the first quadrant, (2) when Pi is in the third and Pj in the fourth quadrant, (3) when Pi is in the fourth and P* in the second quadrant. 6. Find the lengths of the sides of the following triangles: (1) (2, 3), (-5, 8), (-2, -4). (2) (3, -6), (0, 5), (-4, -2). 6. Show that the points (9, 12), (-3, -4), and (5, 4 - 4\/§) lie on a circle whose center is at the point (3, 4). 7. Find a point whose abscissa is 3 and whose distance from (—3, 6) is 10. ^ (**.»♦) Fig. 12. Digitized by Google 520] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 19 Suggestion. — Let y be the ordinate of the point. Then V(3 + 3)* + (y - 6)» - 10. Solve for y. 8. Find the center of the circle passing through the three points (6, 15), (13, 8), and (-4, -9). Suggestion. — By the definition of a circle, if a circle passes through these three points, there must be a point (x, y) from which they are equally distant. Write the distance of each point from the point (x, y) and form two equations. Solve these equations for x and,y. 9. Three vertices of a parallelogram are (—2, 4), (5, 2), and (6, 1). Find a fourth vertex. How many are there? Suggestion. — Use the fact that the opposite sides of a parallelogram are equal. i 10. Two vertices of an equilateral triangle are (2, 10) and (8, 2). Find the third vertex. 11. Find the equation which states that the point (x, y) is 5 units from the point (3, 4). What is the locus of the point (x, y)? Draw the locus. 12. Find the equation that expresses the fact that the point (x, y) is equally distant from the points (2, 3) and (7, —4). What is the locus? 13. Show that the values of line segments parallel to either axes in rectangular coordinates hold true when the axes are oblique. 14. If the axes are inclined to each other at an angle of <•>, and if lines PiQ and QP t of Fig. 11 are drawn parallel to the axes, then the angle P\QP% equals <•> or 180° — w. By the cosine theorem of t igonom- etry show that then the distance between the two points Pi(x\, yQ and P«(*i, Vt) is d = V(*i - xt) 1 + (yi - y t )* + 2{x x - x t )(yi - y%) cos «. 16. The angle between two oblique axes is 60°. Find the distance between the points (—2, 3) and (6, —4). DIVISION OF A LINE SEGMENT 20. Internal and external division of a line segment. — If Pi and P% are any two points on a straight line, then any third point, P , on the line is said to divide the line seg- -& ? l ^° ?* p ° H h- ment PiP 2 into two parts. Fjq 13 The point P is said to divide the line segment PiP 2 internally if P lies between Pi and P 2 ; and externally if P lies beyond P 2 as at P , or be- yond Pi as at Pj. Digitized by Google 20 ANALYTIC GEOMETRY [§21 When P lies between Pi and P* the two parts are P1P0 and PoPj. When P lies at Pq beyond P 2 , the two parts are P\Po and PoPj. When P lies at P* beyond Pi, the two parts are P1P0 and P0P2. The parts are always read as here, that is, from the initial point to the division point and from the division point to the terminal point. When the line segment P1P2 is divided internally, both P1P0 and P0P2 are read in the same direction, and therefore P P the ratio p p is positive, and has a small value when P is * 0* 2 near Pi, and a large value when P is near P 2 , that is, the value of the ratio is between and + » . When the line segment P1P2 is divided externally the two parts are read in opposite directions, and therefore the ratio P P f P P w WWf or BTjd°, is negative. Further, when the point of '0*2 *0* 2 division lies beyond P% the ratio is between — » and — 1, and when the point of division lies beyond Pi the ratio is between — 1 and 0. It remains to express these geometric ideas analytically. This is done in the next articles. EXERCISES 1. Upon a straight line locate two points Pi and P* 6 units apart. P P P P Locate a third point P such that ^^ = §. Such that —^ = — f . Suggestion. — These may be determined by methods of plane geometry, or may be computed by algebra. 2. Divide a line 4 in. long into two parts that are in the ratio 3:1. In the ratio —5. 21. To find the coordinates of a point that divides a line segment in a given ratio. Example 1. Internal paint — Required the coordinates of the point that divides the line segment from Pi( — 2, —4) to P*(5, 6) in the ratio |. Solution. — Draw a pair of axes as in Fig. 14, and locate the points Pi and Pi. Let Po(x , y ) be the required point. Draw lines through these points parallel to the y-axis_ and cutting the x-axis in Afi, Mi, and Mo respectively. Digitized by Google §21] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 21 Then, by plane geometry, ' . / Mom s PiPo P0P1 p.(M) **X But M iM o — «o — ( — 2) and MoJlf » - 5 — So, by [2J, for the abscissas of Af i, Af , and Mt are respectively —2, a? , and 5. And it is given that p^ — 5. Honflfl a? - (-2) 5 Hence 5-Xo = 2' Solving this equation, jr » 3. Similarly, draw lines parallel to the s-axis cutting the y-axis in Ni iVo, and Nt respectively. TKan AW PiP Then m; " PoK But N1N0 - yo - (-4) and ATotfi - 6 - yo. [2,] Hence ^=-^4 ^ o — Vo * Solving this equation, y = 3-^. Therefore the point P has as coordinates (3, 3\). Example 2. External point. — Required the coordinates of the point that divides the line segment from Pi ( —3, 5) to Pa (2, —3) in the ratio — f . * Solution. — Locate Pi and P* as in Fig. 15. Since the ratio is — £, the point of division, P (zo, yo) must be farther from Pi than from P», and so is beyond P* as shown. Draw the lines PiAf i, P*Af *, and PoM as in example 1. T , M,Mo PiPo 5 inen MoMt = PoP, * 3' But MiMo - xo - (-3) and M M t »' 2-s . [2i]. „ s - (-3) 5 Hence — = — = — «. 2 — x 3 Solving this equation, Xo = 9^. X'-H- t-H-X p o(*o»o) y Fig. 15. Then xr ^ But AT^o - yo - 5 and ATotfi - -3 - y . yo - 5 5 3* Similarly, draw PiiV\, PjiVj, and PqN* Then Tr ^ r = p-p- - - g . [2d. Hence -3 - y Solving this equation, yo ■» — 15. Therefore the point P has as coordinates (9j, —15). Digitized by Google 22 ANALYTIC GEOMETRY [522 X i i l i «Po(*o.»o Example 3. External point. — Required the coordinates of the point that divides the line segment from Pi (5, — 2) to P»( — 2, 4) in the ratio — #. Solution. — Locate Pi and P» as in Fig. 16. Since the ratio is — $, the point of division Pq(x , j/o) must lie nearer to Pi than to P», and so is beyond Pi as shown. Draw the lines PiMi, PjMj, and P Mo as in example 1. ^_ M l Mo PiPo _2 inen M M* = PoP, * 5* But M iAf o = x — 5 and MoJif i - -2 - a*. [2i]. xx 2o — 5 2 Hence — s = —v. —2 — Xq 5 Solving this equation, x — 9$. Similarly, draw PiiVi, PiiVj, and PoiVo. Then iMr; = PoP; = ~s- But NiNo - yo - (-2) and tfoAT, - 4 - y . [2,]. Hence *+* = 4 Solving this equation, y — —6. Therefore the point P has as coordinates (9§, —6). 22. Formulas for finding coordinates of point that divides a line segment in a given ratio. — Required the coordinates of the point that divides the line segment from Pi(xi, yi) to P2O&2, y 2 ) in the ratio ri:r 2 Let Po(x y yo) be the required point. Draw lines through Pi, P 2 , and P parallel to the y-axis and intersecting the z-axis in M\> M 2 , and M respectively. M1M0 _ P1P0 _n r% [2J Then M0M2 ~~ P0P2 But M 1M0 = x — xi and M<Mi = x* — x . Xo "" Xl f*i £2 ~~ x <> r i Solving for x 0} x = * 2 , — — • ri + r a Similarly, draw lines through Pi, P 2 , and P parallel to Hence Digitized by Google §22] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 23 the x-axis and intersecting the y-axis in N h iV a , and iV respectively. tv^ N * N * _ PJP± _ ri inen iV~o# 2 " P0P2 " r,' But iViiVo = 2/0 — 2/1 and iWVi = 2/2-00. [2*] Hence *^» = *• 2/2 - 2/0 r 2 Solvingforyo, »» - r ? + l* 1 ri •+■ r 2 Therefore the coordinates of P are rAi ~ - ^1X2 + r 2 Xi riy 2 + rtyi ri + r 2 ri + r 2 Special case. — It is frequently required to find the coordi- nates of the point bisecting a line segment. In this case the two parts are equal, and the ratio — = 1. Formula [4] then becomes [6] x, = *+*, yo =y±±i>. It is readily seen that the results of the last two articles are true for oblique axes as well as for rectangular axes. EXERCISES In the first four exercises draw the figure, and solve without using formulas [4] and [6]. 1. Find the coordinates of the point which divides the line from (-5, -8) to (-1, 4) in the ratio 3 : 1. 2. Find the coordinates of the point which divides the line from (-1, 4) to (8, 1) in the ratio 1:3. 3. Find the coordinates of the point which bisects the line from (8, 6) to (-2/ -3). 4. Find the coordinates of the point which divides the line from (-4, 8) to (2, 6) in the ratio -|. 6. Do each of the first four exercises by the formulas. 6. Find the coordinates of the point which divides the line from (3, -9) to (-1, 5) in the ratio 5:3. 7. Find the coordinates of the point which divides the line from (-6, 8) to (3, -2) in the ratio 3:1. In the ratio -2:3. Digitized by Google 24 ANALYTIC GEOMETRY [$23 8. Draw a triangle the coordinates of whose vertices are (1, 1), (2, —3), and (—4, —6), and find the coordinates of the middle points of its sides. 9. The coordinates of P are (2, 3) and of Q are (3, 4). Find the coordinates of R so that PR : RQ - 3 : 4. 10. Draw the triangle with vertices at (3, 5), (—5, —3), and (9, —7). Find the lengths of its medians. 11. Show that the line joining the middle points of two sides of the triangle, having as vertices the points (8, 6), (1, 1), and (4, —5), is equal to one-half the third side. 12. Prove that the diagonals of the parallelogram whose vertices are (10, 4), (—3, 4), (-6, -6), and (7, -6) bisect each other. 13. The middle point of a line is at (4, 6) and one extremity is at (—3,-2). Find the other extremity. 14. Find the coordinates of the points that trisect the line from (2, 2) to (-7, -4). 16. Show that the median of the trapezoid whose vertices are (0, 0), (a, 0), (6, c), and (d, c) equals one-half the sum of the parallel sides. ANGLES FORMED BY LINES 23. The angle between two lines. — That one line forms an angle with another is a geo- metric idea, and does not neces- sarily depend upon whether or not the lines are considered as having a positive or negative sense, that is, direction. In order to express the facts analyti- cally, we start with the following : Definition. — The angle that a line h makes with a line l* is the angle, not greater than 180°, generated by revolving U in a positive direction until it coincides with h. In Fig. 17, both (a) and (fc), the angle <p is the angle that h makes with 1%. It follows that the angle that h makes with h is t — (p. The definition still holds when the lines do not intersect, that is, are not in the same plane, if it is understood that 1% revolves in a plane parallel to l u until it is parallel to h. Digitized by Google §24] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 25 24. Inclination and slope of a line. — An important special case of the angle that one line makes with another is the angle that a line makes with the x-axis. This angle is called the inclination of the line. It is always measured from the positive part of the x-axis. Thus, in Fig. 18, a\ is the inclination of h v and a s the inclination of 1%. ^ +-X Fio. 18. Definition. — The tangent of the inclination of a line is called the slope of the line. Thus, if to is the slope of a line and a its inclination, then m » tan a. Since the inclination may be any angle in the first or second quadrant, the slope of a line may have any real value either positive or negative, including and ± « . 25. Analytical expression for slope of a line. Example. — Required the slope and inclination of a line I passing through the points Pi (2, 3) and P»(5, 6). Solution. — Locate the points Pi and P% and draw the line J as in Fig. 19. Through Pi draw a line parallel to the x-axis, and through Pt a line parallel to the y-axis. These lines meet at Q, and the angle QP\P% is the inclination of I. Then to = tan QP\P%. 6-3 But tan QPiP, - ^ 5-2 tan" 1 ! - 1. 45°. Fig. 19. Hence to =» 1, and a It should be noted that, whatever the position of the points, the line drawn parallel to the x-axis is so drawn that an angle equal to the inclination is formed. 26. Formula for finding the slope of a line through two points. — Required to find the slope of a line I in terms of the coordinates of two points P\(x\, yi) and P^x*, y 2 ) on the line. Let the line be in either of the positions shown in Fig. 20. In either case draw a line through Pi parallel to the rc-axis, Digitized by Google 26 ANALYTIC GEOMETRY [§27 Pi<Vi)f r i<Vt) forming an angle equal to a as shown; and through P 2 draw a line parallel to the y-axis meeting the first line in Q. Then, whether the slope m of I is positive or negative, m = tan a = ~~rk = PiQ x 2 — xi If Pi and P* are inter- changed, the slope is — — — , X\ £j which equals Vt ~~ Vl - H X t - Xj Therefore, in any case, the formula is [6] m - tan a = iLZJl. Xi — x 2 27. The tangent of [the angle that one line makes with another in terms of their slopes. — Required the tangent of the angle that line It, having a slope of m h makes with 1%, having a slope of m*. Let the inclinations of h and h be ct\ and a* respectively. Pi(*i.»> t4 \+X I Fio. 20. +-X Fig. 21. *-X Then tan a\ = m h and tan a* = m*. There are two cases: case I when ai>a 2 , Fig. 21 (a); and case II when ai<a%, Fig. 21 (6). In each case, let <p be the angle that U makes with 1%.' Then, in case I, a\ = as + ^>, or <p = ai — as. And, in case II, a 2 = a\ + (180° — ^>), or <p = 180° + (ai - a*). Digitized by Google §28] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 27 In either case, t . tan a x - tan a, m x - m 2 tan ^ = tan (ai — a*) = — [7] tan =*p 1 + tan ai tan a 2 1 + mim 2 nil — mi 1 + mim* 28. Parallel and perpendicular lines. — // two lines are parallel, their slopes are equal, and conversely. If two lines are perpendicular to each other, the slope of one is the negative reciprocal of the slope of the other, and conversely. If line* h is parallel to line U, then ai = at, and mi = m%. Conversely. If m, = tn t , ?\~ m * = 0. Then tan <p = 0, and hence <p = 0. Therefore h and 1% are parallel by Art 23, If ii and U are perpendicular to each other, ai = a%+ 90°, or ai = a t - 90°. Then tan ai = tan (at + 90°) = — cot at = — t tan at Therefore m\ = , and m 2 = f?l 2 Wtl Conversely. If mi = , tan ai = — =* — cot at. * m 2 tan a 2 But cot at = tan (90° - a 2 ) = -tan (a 2 - 90°), or, cot a 2 = -tan (90° + a*). Then tan ai = tan (a 2 - 90°), or tan ai = tan (90° + a*). From this ai = a 2 - 90°, or ai = 90° + a 2 . Hence either a 2 — ai = 90° or aj — a 2 = 90°. Therefore <p = 90°, and Zi and 1% are perpendicular to each other. The following are the important facts to remember: [8] For parallel lines, mi = m 2 . [9] For perpendicular lines, mi = , and m 2 = m 2 mi Example. — Find the angle that the line through (4, 5) and (—2, —4) makes with the line through (0, 4) and (—6, —8). Digitized by Google *p 1 +1-2 - (-}) = tan" 1 (-4) - 172° 52.4'. 28 ANALYTIC GEOMETRY [§28 Solution. — The slope of 1 1 is mi — t~t~o " 5* 4+8 The slope of /» is m% - r T — 2. o •+• o 1 — 2 1 Substituting in [4], tan <p — i J. i.g ™ "~ S* r 2°l EXERCISES 1. Find the slopes of the lines through the following pairs of points: (1) (-4, -4) and (4, 4). (4) (-V5, V5> and (\/2, V^). (2) (-4, 3) and (-3, 2). (5) (-a, 6) and (c, d). (3) (5, 0) and (6, VS). (6) (\/3, 2) and (\/2, 3). 2. Find the inclination of each of the lines of exercise 1. 3. Find the slope of a line that is perpendicular to the line through the points (3, 4) and (-2, -3). 4. Show that the line through (4, 2) and (3, 7) is perpendicular to the line through (8, 1) and (13, 2). 5. Find the value of y so that the line through (3, 7) and (4, y) shall be perpendicular to the line through (9, 10) and (6, 8). 6. Prove by means of slopes that the three points (6, —3), (2, 3), and (—2, 9) are on the same straight line. 7. Find the value of x so that the three points (x, 6), (2, 8), and (4, 7) shall be on the same straight line. 8. Express by an equation the fact that a line passing through the points (4, 5) and Or, y) has a slope of f. 9. A line passes through the point (—4, 6) and has a slope of — f. Find the abscissa of the point on the line whose ordinate is —3. 10. Express by an equation the fact that a line passing through the point (—3, —6) is perpendicular to the line through the points (—2, 7) and (4, 6). 11. Express by an equation the fact that a line passing through the point (7, 2) is parallel to the line through (—6, —2) and (4, —7). Find the point on this line whose abscissa is —3. 12. Two lines l\ and U make tan~4 and tan _1 ( — !) respectively with the x-axis. Find the angle that l\ makes with U. 13. Find the slope of the line that makes an angle of 47° with the line having a slope of 0.3674. Suggestion. — Substitute <p «■ 47° and m 8 =* 0.3674 in [7] and solve for mi. 14. Find the angle that the line through the points (—3, 6) and (4, —2) makes with the line through the points (1, 1) and (—7, —7). Digitized by Google §29] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 29 15. A line passes through the point (4, 5) and is parallel to the line through the points (3, -2) and (-2, 5). Find where the line cuts the y-axis. 16. A line I makes an angle of 30° with the line through the points (2, 3) and (6, 7). Find the slope of I. 17. Show that the lines joining the middle points of the sides of a quadrilateral whose vertices are the points (5, —4), (3, 6), (— 1, 4), and (—3, —2) taken in order, form a parallelogram. 18. Prove by means of the slopes of the sides that the quadrilateral whose vertices are the points (4, 2), (2, 6), (6, 8), and (8, 4) is a rectangle. 19. A point is equidistant from the points (—5, —2) and (2, —5), and the line joining the point to (4, 2) has a slope of — J. Find the codrdi- nates of the point. 20. A line passes through the point (4, 5) and has a slope of 0.7236. Find the ordinate of the point on this line having as abscissa —2. 21/ The vertices of a triangle are Pi(3, 4), P*(-4, 3), and Pi(-1, -4). Find the angle of the triangle at the vertex P t . POLAR COORDINATES 29. Location of points in a plane by polar coordinates. — Thus far only the first method mentioned in article 13 for PlP.O) Fig. 22. locating points, has been used. The second method, that by polar coordinates, has advantages over the cartesian system in certain oases. This method will now be explained. In polar coordinates we locate a point in a plane by giving its distance and direction from a given fixed point in the plane. Thus, in Fig. 22, given the fixed point in the fixed directed line OX, then any point P in the plane may be located by stating its distance OP = p from 0, and the angle through which OX must turn to coincide with OP. . Definiti<ms^--The fixed point is called the pole or origins Digitizedby Google 30 ANALYTIC GEOMETRY [§29 the fixed line OX the initial line, or polar axis ; the line segment OP = p is called the radius vector of P; and the angle 6 the vectorial or directional angle of P. Together, p and $ are the polar coordinates of P, and are written (p, 6). P X {W) -**x P* (V30 ) Fig. 23. iP.C-i.-ViT) A(-a.30°) Fio. 24. In order to use both positive and negative numbers as coordinates of points, the usual conventions of trigonometry as to positive and negative angles of any size are accepted. It is also agreed that the radius vector is positive if measured from along the terminal side of the angle 0, and negative if measured in the opposite direction. Thus, Pi (5,60°) is located as shown in Fig. 23, the angle being mea- sured counter-clockwise and the radius vector along the terminal side in the positive direction. To plot the point P*(4, —30°), the angle is measured clockwise and the radius vector positive. (Fig. 23.) The following points are plotted as shown in Fig. 24: Pi (—3, 30°) andP,(-4, -W. » (-5,240°) >(6,-300 W ) *-x Fig. 25 From the above illustrations it is clear that one pair of polar coordinates determine one, and but one, point in the plane. Digitized by Google §29] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 31 On the other hand, for a single point there are an indefinite number of pairs of polar coordinates. Thus, if only values of 6 numerically less than 360° are taken, then the four pairs of coordinates (5, 60°), (-5, 240°), (-5, -120°), and (5, —300°) all determine the same point as shown in Fig. 25. For convenience in plotting, polar coordinate paper ruled with concentric circles and radial lines, as shown in Fig. 26, can be obtained. The following points are shown plotted in 105° 90° »co m Fio. 26. Fig. 26: P(5, 20°), Q(-6, 80°), fi(8, fcr), S(-7, fir), and T(-8, -fr). EXERCISES 1. Plot the following points in polar coordinates: (1) (3,30°). (6) (3, -**). (11) (3,1). (2) (7, 120°). (7) (-4,*). (12) (-4, -2). (3) (-2,40°). (8) (-2, -»). (13) (5, -3). (4) (-6, 150°). (9) (2, 0). (14) (-6, -5). (5) (4, -75°). (10) (-6 fr). (15) Or, -») 2. Give three other pairs of coordinates in which is numerically less than 360° for each of the following points: (1) (7, 30°), (2) (-3, |x). Digitized by Google 32 ANALYTIC GEOMETRY [§30 3. The side of a square is 4 in. and the diagonal is taken as the polar axis with the pole at a vertex. Find the polar coordinates of the vertices. 4. What is the locus of all points for which p — 5? For which $ = $*-? For which - Jir? 30. Relations between rectangular and polar coordi- nates. — Let X'X and Y'Y, Fig. 27, be a set of rectangular coordinate axes; and let the polar axis OX be taken on the positive part of the s-axis with the pole at the origin. Let P be any point in the plane. Draw OP y and QP per- pendicular to X'X. Then by the definitions already given, OQ = x, QP = y, OP = p, and L XOP = 6. .r By trigonometry and geometry it follows that * i P x ' V V V . \ r 9 r' x < I [10] x = o cos 0, y = p sin 0, x* + y* = p 2 . By means of these formulas polar coordinates can be ex- pressed in rectangular codrdi- Fl °- 27 ' nates. Also by trigonometry and geometry it follows that [11] q - V* 2 + y 2 , X 6 = tan- By means of these formulas rectangular coordinates can be expressed in polar coordinates. EXERCISES 1. The origin in rectangular coordinates coincides with the pole in the polar system, and the x-axis falls upon the polar axis. Find the rectangular coordinates of the following points: (6, !*■), (—2, !*■), (-5, M, (6, W, (3, fr), (8, ix), (2, *), (6, fr). 2. Find the rectangular coordinates of the points whose polar coordi- nates are: (2, 40°), (3, 70°), (6.5, -30°), (1.2, 130°), (-4.5, 155°). Digitized by Google §31] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 33 3. Find two pairs of polar coordinates for each of the following: (4, 4V3), (-3, -3), (3, 5), (-y% VS). 4. By means of [10] derive d « Vpi* + pi* — 2pipi cos (0i — 0») from [3]. Here (xi, y\) and (pi, 0i) are the same point, as also are {x%, y t ) and (j>t, 0*). 5. Derive the formula for the distance between two points in polar coordinates directly by means of the cosine theorem of trigonometry. 6. The polar coordinates of P are (5, 75°) and of Q are (4, 15°). Find the distance PQ. TRANSFORMATION OF RECTANGULAR COORDINATES 31. Changing from one system of axes to another. — From the discussions already given, it is evident that the coordinate axes may be chosen at pleasure. In any particular case it is clear that they should be so chosen that they can be used to the best advantage. In order to discuss certain problems that occur in analytic geometry, it is necessary to express the coordinates of points in the plane in another system of coordinates than that in whifeh they are already expressed. It is of advantage to deduce formulas for making these transformations which are of two kinds; (1) Transformation by trans- lation of axes, or changing to new axes that are parallel to the original axes. ,(2) Transformation by rotation' of axes, or changing to new axes that make a certain angle with a the original axes. 32. Translation of coordinate axes.— Let OX and OY, Fig. 28, be any system of cartesian axes; and let O'X' and O'Y' be another set parallel to the original. Let 0', the origin of the new system, have coordinates (h, k) when referred to the original system. N Y Y' i f«:*') B & 1 O A M ' Fio. 28. Digitized by Google 34 ANALYTIC GEOMETRY [§33 Let P be any point in the plane, having coordinates (x, y) when referred to the original system and (a/, y') when referred to the new system. Draw a line through P parallel to the x-axis, intersecting OY in N and O'Y' in N'. Also draw a line through P parallel to the y-axis, intersecting OX in M and O'X' in M'. Then NP - 2W + tf'P, and AfP = MM' + M'P. Arts. 10 and 18. But NP - x, iW = A, tf'P - x 7 , MP - y, MM' = fc, M'P = y'. Therefore the formulas for translating the axes are: [12] x = if + h, y = y' + k. • Solving these formulas for x' and y', [12 J * - x - h, y' - y - k. In this article it is not implied that the axes are rectangular, and therefore the formulas hold for transforming from any set of cartesian coordinate axes to a parallel set. 33. Rotation of axes. Transformation to axes making an angle <p with the original.— Let imW) OX and Y, Fig. 29, be any system rX f of rectangular axes, and let OX' and OY 9 be another set of rec- tangular axes having the same origin as the original, but making an angle <p with OX and OY respectively. Let P be any point in the plane, having coordinates (x, y) when re- ferred to the original system, and (a/, y') when referred to the new system of axes. Join to P, draw MP perpendicular to OX, draw M'P perpendicular to OX'. Let LXOP = 6, and /.X'OP = 0', and OP = p. +>x Digitized by Google §33] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 35 Then x = p cos 6 = p cos (0' + <p) = p cos $' cos <p — p sin 0' sin <p. [10] And y = p sin0 = p sin (0' + <p) = p sin 0' cos ^ + p cos 0' sin <p. [10] But x 7 = p cos 0' and y' ■» p sin 0'. [10] Substituting these values, [13] x = x' cos tp — y' sin ^, y = x' sin ? + y* cos ^. Solving these formulas for x' and y', [13i] *' = x cos <f> + y sin ?, y' = y cos ^ — x sin <p. Example.— The point P has coordinates (\/2; 2-\/2) when refeixed to a certain system of rectangular axes. Find the coordinates of P when referred to a new set of rectangular axes having the same origin but making an angle of 45° with the original. ! Solution. — Here we are to find x' and y' when x, y, and <p are known, and so we use formulas [13 J. Substituting in these formulas, x' - V2 cos 45° + 2\/2 sin 45° - V2 X iV2 + 2V2 X i V2 - 1 + 2 - % I/' = 2\/2 cos 45° - V2 sin 45° - 2V2 X i\/2 - V2 X i\/2 =2-1-1. Check the values by a drawing. EXERCISES 1. Find the coordinates of the following points when referred to axes parallel to the original and with origin at the point (3, 4): (7, 8), (4, 3), (0,0), (-2,6), (-7, -5), (6, -8). 2. Find the coordinates of the following points when referred to axes having the same origin as the original, but making an angle of 45° with them: (2, 3), (-3, 4), (-5, -5), (7, -1). 3. The coordinates of the vertices of a triangle are Pi (—3, —4), Pi(6, —2), and P*(2, 7). Find the coordinates of the vertices when referred to parallel axes with origin at Pi. Plot. 4. The coordinates of a point Pi are (3, 2). Find the coordinates of the origin of a new set of axes parallel to the old so that the coordi- nates of Pi shall be (—4, —6) when referred to the new axes. 5. The coordinates of the vertices of a triangle are Pi(0, 0), P»(2, 2\/3)» Digitized by Google 36 ANALYTIC GEOMETRY [§34 and P|( — 2, 4). Find the angle through which the axes must be rotated so that Pj shall lie on the new x-axis. Find the coordinates of the vertices of the triangle referred to the new axes. 6. Derive formulas [13i] from Fig. 29 without solving [IS]. • AREAS OP POLYGONS 34. Area of a triangle in rectangular coordinates. — Let APiPjPa, Fig. 30, be any triangle having vertices Pi(xi,yi), P*(x 2 , y % ), and P 8 (x 8 , yz) re- Af«fclTi) ferred to the axes OX and OY. Translate the axes to a new system having as origin one p 8 (**¥,)vertex of the triangle, say Pi (x h yi), Then the codrdinates of P 2 and P 8 referred to O'X' and O'Y' are P 2 (x 2 ', y 2 ') and Pi{x*, yz) respectively, where Xi' = x t - xi, yj = yi - yi, ^x' *»x Xi = Xz zi> V* = y» - yi Fio. 30. by[12J. Let ZX'O'P* = 2 , and /.X'O'Pz = 8 . The area of APiPjP* = \PiP% times the altitude from P a to PiP 2 . Hence area APiPaPa = \PiP% X PiP 8 sin (0 8 - 2 ) — \P\P% X PiPs (sin $z cos 2 - cos $z sin 2 ) = KP1P2 cos 2 X PiP 8 sin0 8 -PiPs cos 8 X P1P2 sin 2 ). ButPiP 2 cos 2 = x 2 ' = x% — Xi, PiPs sin 0* = y 8 ' = y z — 1/1, P1P3 cos 8 = x 8 ' = x 8 — Xi, PiP 2 sin 2 = y 2 ' = y 2 — y x . Substituting these values and putting A for area of APiP^s, ^ = MO** - *0 (2/3 - yi) - (x 8 - Xi) (yi - 2/1)]. Multiplying and arranging, [14] A = £(xijr« - xjyi + x^ s - x 8 y 2 + x 8 yi - xiy 8 ). Digitized by Google §34] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 37 This may be written in the determinant form Xi 2/1 1 A = \ x 2 2/2 1 , x»y z l by which it can be readily remembered. The formula can also be remembered and the computation carried out by the following: Rule. — First, write down in a line the abscissas of the vertices taken in counter-clockwise order, repeating the first abscissa at the end; and under the abscissas write the cor- responding ordinates. Thus, x\ Xi xz x\ 2/i 2/2 2/8 2/i Second, multiply each abscissa by the ordinate of the following column, and add. This gives xty* + x*yz +x 8 2/i. Third, multiply each ordinate by the abscissa of the follow- ing column and add. This gives yyx* + y&i + yzXi. Fourth, subtract the third from the second and divide by 2. This gives the area as in formula [14]. It is evident that the expression \P x Pt X PiPz sin (0 3 - 0a) for the area is positive or negative according as sin (03 — 02) is positive or negative. In order then to have the area positive, sin (08 — 02) must be positive. Hence 03 — 02 must be positive, and 03>02. That is, P1P2 is turned counter- clockwise to coincide with P1P3. This will be true only if, in passing around the triangle in the order the vertices are taken, the area is always at the left as shown in Fig. 31. That is, a point moving around the triangle must move counterclockwise. Otherwise the area will be negative. Thus, in Fig. 31, the area of the triangle, if the vertices are taken in the order Pi, P%, P it is A - i[3(-5) - 9(-2) + 9-4 - 10(-5) + 10(-2) - 3-4] = 28i. P,(io.O +~x P,(9r5) Fig. 31. Digitized by Google 38 ANALYTIC GEOMETRY [§35 If, however, the vertices are taken in the order P lt P%, P», the area is A - J[3-4 - 10(-2) + 10(-5) - 9-4 + 9(-2) - 3(-5)] = -28*. 35. Area of any polygon. — Any polygon having its verticfes given in rectangular coordinates can be divided into triangles by diagonals drawn from any vertex. Its area can then be found. It can be readily shown that the area of any polygon can be found by the rule given for finding the area of a triangle. Thus, a polygon, Fig. 32, having five vertices as follows, ,taken in order counter-clockwise; Pi(xi,yi) 9 P%(xt 9 2/2), P*(xz, yz) Pa(x4, 2/4), and P 6 (x 6y 2/5), has its v area represented by the scheme, X\ x% x$ x& av#i y\ y% y* y\ 9*fo which evaluated by the rule gives -A = |[(xi2/2 + xjyi'+ x z y 4 + x 4 y b + x 5 2/i) - (2/1X2 + 2/2X3 + 2/3X4 + 2/4X5 + 2/5X1)]. Example. — Find the area of the polygon having the following vertices: (—3, 6), (2,-4), (8,1), and (4, 7). Solution. — Arranging the abscissas and ordinates, -3 2^4 6-417 A - §U(-3)(-4) + 2 1 + 8-7 + 4-6] - [6-2 + ( -4)8 + 1-4 + 7( -3)1 } = 65* square units. EXERCISES 1 in each case draw the figure: (1) (0,0), (10, 12), (-6,8). (2) (-4,6), (-2,9), (10, -4). (3) (17,2), (-3,9), (-6, -10) (4) (0,7), (10, -3), (-4,9). -3 6 Fig. 33. Find the area of the triangles having the following points as vertices, Digitized by Google §36] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 39 2. Find the area of the quadrilateral with vertices (2, 5), (—7, 9), (-10, -3), and (6, -9). 3. Find the area of the pentagon with vertices (1, — 2), (3, 1),(6, 2), (4, -4), and (2, -5). 4. Show that the area of the triangle with vertices (2, 4), (— 6, — 8), * and (1#, —4) is four times the area of the triangle formed by joining the middle points of the sides. 5. Find the area of the isosceles triangle with vertices (4, 5), (10, 13), and (4, 15). Find the altitude from the vertex at (4, 5), and find the' area as one-half the product of the base and altitude. Do the two results agree? 6. Find the area of the triangle with vertices Pi(7, 9), P»(— 6, -8), and Pi (4, —6). Find the point P 4 dividing PjPj in the ratio 2:3, and show by areas that the triangle is divided into two triangles the areas •f which are in the ratio 2:3. 7. If the vertices of a triangle in polar coordinates are Pi (pi, 0i), Pj(ps, #*), and P t (p if # t ), derive a formula for its area. Suggestion. — In [14] put Xi = pi cos $ h y x — p x sin e lf and similarly for P s and P s . Arrange and apply the subtraction formula for sines. 8. Find the area of a triangle the vertices of which in polar coordinates are (10, 30°), (-12, 120°), and (6, 135°). APPLICATIONS 36. Analytic methods applied to the proofs of geometric theorems. — One of the necessary conditions for the mastery of a mathematical subject is a thorough understanding of the fundamental ideas and methods. In the present chapter, stress has been laid upon the expressing of geometric ideas in an analytic form. Time will be well spent in reviewing these methods until they are fully comprehended. As will be repeatedly found in Subsequent chapters, analytic geometry gives a powerful method for treating a great variety of geometric questions. As an illustration of this a few elementary examples of the application of algebra to geometry are given in this article. One of the great advantages of the analytic method of solving geometric problems lies in the fact that an analytic result obtained by the simplest arrangement of the axes with Digitized by Google 40 ANALYTIC GEOMETRY [536 Pt(*t.*t) reference to the geometric figure holds equally well for all other arrangements of the axes. It is well then always to make use of the simplest relations between the geometric figure and the coordinate axes. Example 1. — Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to one-half the third side and is parallel to it. Given any triangle OPiPt, and AB bisect- ing OP* and PiP,. To prove AB = \OP u and that AB is parallel to OPi. Proof. — Choose the coordinate axes with origin at and the 2-axis along OPi. Then the coordinates of are (0, 0), and Pi and Pi may be designated by (xi, 0) and (x», y») respectively. Codrdinatesof A andBare fe ^\ and fo + Z \ ^\ respectively. [5] length of AB - VH^ 4) '+(§-!"' = * 9] PxlmiJh* But OPx - xi - - Xi. . AB = iOPi. [6] P%t*%*4 Also slope of AB = 0, and slope of OP x =* 0, .\ AB is parallel to OPi. To see the desirability of this choice of the axes, the student should write out the proof when the vertices of the triangle are (x h yi), (xi, y%), and (x h yi). Example 2. — Derive a formula for the center of gravity of a triangle with vertices Pi(xi, yi), Pt(xt, Vt), and Pi(x«, y*); it being known that the center of gravity of a triangle is at the intersection "" of its medians, which is two-thirds of the length of any median from a vertex. Solution. — Here no choice of axes can be made that will simplify the work. *•<•** Pll*h*l) Fig. 35. Digitized by Google §36] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 41 Choose any median as P\Q. Then it is required to find the coordinates (*o, Vo) of P such that PiP : PoQ -2:1. Codrdinates of Q are £* + *, UL ^ Jl ) ' By[ 4] ^- ^ + ; +ap s «nd^- y '+;+^ EXERCISES 1. Use the formulas derived in example 2 and find the codrdinates of the center of gravity of a triangle with vertices (2, 6), (-8, 3), and (-4, -3). Prove the theorems in exercises 2-12. 2. The diagonals of a rectangle are equal. 3. The diagonals of a parallelogram bisect each other. 4. The medians of a triangle intersect in a point which is two-thirds the length of any median from a vertex. 6. The middle point of the hypotenuse of a right triangle is equally distant from the three vertices. 6. The diagonals of a square are perpendicular to each other. 7. If the diagonals of a parallelogram are equal, the figure is a rectangle. 8. The distance between the middle points of the non-parallel sides of a trapezoid is equal to half the sum of the parallel sides. 9. The lines joining the middle points of the successive sides of any quadrilateral form a parallelogram. 10. The lines joining the middle points of the successive sides of any rectangle form a rhombus. 11. In any quadrilateral, the lines joining the middle points of the opposite sides, and the line joining the middle points of the diagonals meet in a point and bisect each other. 12. The sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of its diagonals. 13. Given Pi any point in the plane of a rectangle, prove that the sum of the squares of the distances from Pi to two opposite vertices of the rectangle is equal to the sum of the squares of the distances from Pi to the other two vertices. GENERAL EXERCISES 1. If the points A, B, C, D, and E are any points on the same straight line, show that: Digitized by Google 42 ANALYTIC GEOMETRY [§36 (1) AB +CD -CB -ED - AE. (2) AE + EB + DE + EC - DB - ilC. (3) AC - J£B + CB - AE - 0. 2. If the coordinates of the vertices of a rectangle are (0, 0), (8, 0), (8, 6), and (0, 6), what will be the oblique coordinates of its vertices if the ff-axis is the diagonal through the origin, the a>axis remaining as before? 3. What are the oblique coordinates of the vertices of the rectangle of exercise 2 if the y-Qjda is taken as the diagonal through the point (0, 6)? 4. What are the coordinates of the vertices of a square if a side is 4\/2. and its diagonals are taken as the coordinate axes? 6. A rhombus lies wholly in the first quadrant, and the angle between two of its sides is 30°. If the coordinates of two of its vertices are (0, 0) and (a, 0), find the coordinates of the remaining vertices. 6. Find the coordinates of the vertices of an equilateral triangle of side a if its center is at the origin and the y-axis passes through one vertex. 7. The angle between two oblique axes is 135°. Find the distance between the points (1, 3) and ( — 1, —3). 8. What is the ratio in which the j/-axis divides the line segment joining (-2, 3) to (5, -1)? 9. Find the coordinates of two points which divide the line segment from (2, 4) to (8, —8) internally and externally in the ratio whose numerical value is 2. 10. find the coordinates of Pi and P* where Pi is on the positive j/-axis, Pi on the positive x-axis, and the point (2, 3) divides PiPj in the ratio 2:1. 11. The point (-2, -2) divides the line PiPj in the ratio -4:3. If Pi has the codrdinates (2, 6), find the coordinates of P*. 12. If Pi has the codrdinates (1, 4) and P s the codrdinates (5, 1), find a point P 8 on P1P2 such that PiPj will be a mean proportional between PiPs and 25. 13. Prove analytically that the diagonals of a rhombus intersect at right angles. 14. The hypotenuse of a right triangle is the line joining ( — 1, —2) to (6, 4). Find the coordinates of the third vertex if it lies on the x-axis. 16. One end of the line whose length is 5 is at (4, 2). The abscissa of the other end is 7, what is its ordinate? 16. The end points of a diagonal of a parallelogram are (2, —3) and (3, 2). Find the codrdinates of the remaining vertices if they are on the x-axis and y-axis respectively. Why is there only one solution? Digitized by Google §36] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 43 17. The codrdinates of the end points of one diagonal of a rhombus are (0, 0) and (2, 4). If one side lies along the positive x-axis, find the codrdinates of the end points of the other diagonal. 18. Two points Pi and P% are at the same distance from the origin. If their polar codrdinates are (p, 0i) and (p, 0*), show that the slope of the line joining them is —cot — - — • 19. What does the slope of the line joining ( — 1, 3) to (6, 7) become if the axes are rotated through an angle ? » tan" 1 }? 20. What does the slope of the line joining (4, 3) to (—5, 6) become if the axes are rotated through 30°? 21. What is the slope of the line through the points the polar codrdinates of which are (6, 30°) and (4, 60°)? 22. Find the area of a triangle the polar codrdinates of the vertices oft which are (J*-, Jx), (*•, i*-), and (2r, i*-). 23. Find the area of a triangle the polar codrdinates of whose vertices are (1, 60°), (3, 210°), and (2, 240°). 24. A rectangle of sides 5 and 12 lies entirely in the second quadrant, with one vertex at the origin and the longest side on the negative z-axis. Find the codrdinates of its vertices if the axes are revolved so that the y-axis coincides with one diagonal. 25. The codrdinates of the vertices of a parallelogram are (0, 0), (4, —3), (5, 0), and (1, 3). What will be the coordinates of its vertices if the axes are rotated so that the a>axis coincides with the longest side? Digitized by Google CHAPTER III LOCI AND EQUATIONS 37. General statement. — In the present chapter will be considered some of the more simple cases of the first two funda- mental questions of analytic geometry as stated in article 3. The locus of an equation will be considered first, and then the equation of a locus. That is, the geometric interpretation of an equation will be dealt with first. 38. Constants and variables. — A constant is a number that never changes, or one that does not change in the course of a discussion. Constants that never change are definite numbers, as 2, § , \/2, log 3, and x. Numbers that are constant during a dis- cussion, but may be different in another discussion are repre- sented by the letters that are assumed to have known values. A variable is a number whose value changes arbitrarily,* or according to some law. The number expressing the speed of a train as it gains headway is a variable. The price of a stock may change from day to day, and is expressed by a variable. The velocity of a falling body changes from instant to instant, and is expressed by a variable. If two variables are so related that for every value of one there is a corresponding value of the other, then the one is said to be a function of the other. Thus in the formula for the area of a circle, A = irr 2 , for every value of r there is a value of A. Then A is a function of r. This is written A = /(r). Likewise r may be considered a function of A. 44 Digitized by Google $39] ' LOCI AND EQUATIONS 45 EXERCISES 1. Give various illustrations of variables and constants. 2. In the formula, A = $*r», for the volume of a sphere, which are constants and which variables? Is A a function of r? Is r a function of A? 3. In the equation x + y =» 6, can either x or y be assigned values arbitrarily? Can both be given arbitrary values at the same time? Is x a function of y? Is y a function of x? 39. The locus. — If the location of a point is determined by- certain stated conditions, then the locus of the point is the geometric figure such that, (1) all paints of the figure satisfy the given conditions, and (2) all points that satisfy the given conditions are in the figure. In proving that a certain figure is the required locus, it is sometimes more convenient, instead of (2), to prove that any point not in the figure does not satisfy the given conditions. The conditions determining a locus may be stated in the language of geometry, or may be stated by an equation. In the more simple cases the locus can be given immediately from the conditions stated. EXERCISES 1. What is the locus of a point that is equally distant from two fixed points? 2. What is the locus of a point in a plane and at a constant distance from a fixed point in that plane? 3. What is the locus of a point equally distant from two intersecting straight lines and in the plane determined by those lines? 4. What is the locus of a point equally distant from three fixed points and in the plane determined by the three points? 5. In rectangular coordinates, what is the locus of a point whose abscissa is 0? Whose abscissa is 5? Whose ordinate is —6? 6. What is the locus of a point whose coordinates satisfy the equation x = 4? Which satisfy the equation x = y? The equation x + y — 0? 40. The locus of an equation. — If an equation is the analy- tic statement of geometric conditions, then it follows from the definition of a locus. Art 39, that the locus of an equation is Digitized by Google 46 ANALYTIC GEOMETRY [J41 the locus of all points whose coordinates satisfy the equation; and, conversely, the coordinates of all points on the locus must satisfy the equation. While the preceding statement is general, only rectangular coordinates will be used in the present chapter. The drawing of the locus is spoken of as plotting the equa- tion! or plotting the locus of an equation. The locus is called the graph of the equation. 41. Plotting an equation. — The steps in plotting an equation are: (1) Solve the equation for one, or each, variable in terms of the other. (2) Assign convenient values to one variable and determine corres- ponding values of the other vari- able, and arrange in a table. 1 1 iMn 1 I f i » x (3) Choose a suitable unit and plot the pairs of values of the variables. Fig. 36. (4) Connect these points by a smooth curve. The variable to which values are assigned arbitrarily is called the independent variable. The other is then called the dependent variable. Example 1. — Plot the equation 2x + Zy — 13. (1) Solving for y, y = ^ (2) Assign values to x as shown in the following table and determine the corresponding values of y. z -4 -2 -1 1 2 3 5 6i 8 10 y 7 5f 5 4i 31 3 2i 1 -1 -2f (3) Locate a pair of rectangular coordinate axes, choose a suitable unit, and plot the points Pi(-4, 7), P 8 (-2, 5}), Pi(-1, 5), • • • Fig. 36 RA. as shown in Fig. 36. Digitized by Google §42] LOCI AND EQUATIONS 47 (4) Draw a smooth curve through the points. The curve is the locus of the equation and appears to be a straight line. Example 2. — Plot the locus of the equation x* + y % — 4x — 21. (1) Solving for y in terms of x and for x in terms of y, y - ± V21 + 4a; -a: «, and x - 2 ±V25 - y*. From the first it is readily seen that y is imaginary when x< —3 or when x>7, for then 21 + 4x — a?*<0 and the square root is imaginary. likewise, z is imaginary when y < — 5 or when y>5. It is evident then that we should not choose values of x less than — 3 nor greater than 7. And should not choose values of y less than —5 nor greater than 5. (2) Here it is convenient to assign arbitrarily some values to x and some values to y, in each case computing the corresponding values of the other variable. ip ill X y y X ±4.6 | 7 or -3 2 ±5 ±2 6.6 or -2.6 4 ±4.6 : ±4 5 or —1 6 ±3 -2 ±3 ; Fig. 37. (3)' The points are plotted as shown in Fig. 37. (4) A smooth curve is drawn connecting the points. This is the locus of the equation and appears to be a circle. 42. The imaginary number in analytic geometry.— In the plan for plotting numbers in analytic geometry no method is provided for plotting imaginary numbers. It follows then that if one, or both, of the values of the variables satisfying an equation are imaginary or complex numbers, no point can be plotted having these as coordinates. Such numbers are often said to locate imaginary points on a curve. Digitized by Google 48 ANALYTIC GEOMETRY [§43 Some equations, such as x 2 + y 2 = 0, are satisfied by only one pair of real values for x and y. The locus of such an equation is a single point. Thus, x* + y* s= is satisfied only by x = and y = 0. Other equations, such as x 2 + y 2 + 4 = 0, are satisfied by no real values for x and y. The locus of such an equation is wholly imaginary. EXERCISES Plot the loci of the following equations: 1; y = 2x + 4. 12. x* + 3y* = 0. 2. 2x + 3y - 8. 13. s l + y* + 12 - 0. 3. a; - 2y - 6 - 0. 14. x* + 2y* - 8. 4. 3x - 4y - 5 - 0. 15. s l + y l + ftr - 7 5. 16s - 3y - 42. 16. y» = 4* 1 . 6. x* + y* - 25. 17. 9s l - 4y* = 36. 7. x* + y* - 18. 18. 9a* + 4y l - 36. 8. y* = 4a:. 19. 4y l - 9a* - 36. 9. y* - 4a; + 3. 2Q. y = a* 10. x 2 - 8y. 21. y = a* + 3. 11. a* - y* = 4. 22. y = a* - 2s* + 6a; - 3. 23. Plot the following equations upon the same set of axes: (1) x* + y* - 16, (2) x* - y* - 16, (3) y* - x* = 16, (4) -x* - y* = 16. 24. Plot the following equations upon the same set of axes: (l)x« = 2y, (2) x* - -2y, (3) y» = 2s, (4) y* - -2s. DISCUSSION OF EQUATION IN RECTANGULAR COORDINATES 43. Geometric facts from the equation. — Since it is possible to plot but a few points of a curve, the method of determining the curve by points is sufficiently accurate only in the case of simple curves. In general, much help in learn- ing the properties of a curve is gained by a study, or discussion, of the equation. First, it gives exact information regarding the curve; second, it furnishes a test of the accuracy of the plotting; and, third, it usually lessens the labor of plotting. Digitized by Google §44] LOCI AND EQUATIONS 49 The properties of the locus of an equation that can be studied to the best advantage by analytic geometry are the following: (1) The intercepts of the curve. (2) The symmetry of the curve. (3) The extent of the curve. Various other properties can be studied by methods of the calculus. 44. Intercepts. — The x-intercepts of a curve are the abscis- sas of the points where the curve intersects, or meets, the x-axis. The y-intercepts are the ordinates of the points where the curve intersects, or meets, the y-axis. Together the ^intercepts and the ^-intercepts are called the intercepts of the curve. Evidently, the s-intercepts are- found by putting y = in the equation and solving for x. Likewise the y-intercepts are found by putting x = and solving for y. It follows that, if an equation contains no constant term, the curve passes through the origin. Example. — Find the intercepts for the equation 16x* -f 25y* = 400 Putting y - 0, 16x* - 400, or x - ±5. Putting x = 0, 2by % = 400, or y = ±4. .*. the it-intercepts are +5 and —5, and the y-intercepts are +4 and —4. ' EXERCISES Find the intercepts for the following equations: 1. 2x - Zy = 10. 5. 5x*y - 15s + 4y =» 0. 2. x* + y* = 36. 6. y - — -^ ~-^ (*+2)(*-l) 3. 4x* + y* = 64. 7. y* - (* + 2)(* - l)(x - 3). 4. 4x* -f y* - 8s - 2y + 1 - 0. 8. xy - 6. 46. Symmetry, geometrical properties. — Two points are said to be symmetrical with respect to a given point when the given point bisects the line joining the two points. The given point is called the center of symmetry. Two points are said to be symmetrical with respect to a Digitized by Google 50 ANALYTIC GEOMETRY [$46 given line when the given line is the perpendicular bisector of the line joining the two points. The given line is called the axis of symmetry. Thus, in Fig. 38, if Q bisects PiP,, Pi and f s Pi are symmetrical with respect to Q. Also, if A £ is the perpendicular bisector of PiPi, Pi a and Pi are symmetrical with respect to A B. If the points of a curve can be ar- *p f ranged in pairs which are symmetrical Fig. 38 with respect to a line or point, then the curve itself is said to be symmetrical with respect to the line or point. Thus, in Fig. 39, the curve is symmetrical with respect to each of the coordinate axes and with respect to the origin. Tell why. EXERCISES 1. Has a square a center of symmetry? Has a rec- tangle? A circle? A par- ^allelogram? A regular hexagon? 2. How many axes of symmetry has each of the figures of exercise 1? ^ 3. In rectangular coor- dinates give the point that with each of the following is symmetrical with respect to the x-axis: (2, 4), (—2, 5), (—4, —2), (6, —8), (a;, y). With respect to the y-axis. With respect to the origin. 46. Symmetry, algebraic properties.— In the preceding article symmetry has been considered from the side of geometry. It remains to determine how symmetry can be seen by an inspection of the equation. If a curve is symmetrical with respect to the z-axis, it follows that every point (x, y) on the curve has a correspond- ing point (x, —y) on the curve. Then the coordinates of the Digitized by Google §47] LOCI AND EQUATIONS 51 point (x, — y) must satisfy the equation; that is, if — y is substituted for y, the equation reduces to the original form. It is evident that this occurs in an algebraic equation when only even powers of y appear in the equation. (See Art. 120.) Likewise the curve is symmetrical with respect to the y-axis if, when —x is substituted for x, the equation reduces to the original form. This occurs in an algebraic equation when only even powers of x appear in the equation. Since the pair of points (x, y) and (— x, —y) are symmetrical with respect to the origin, it follows that if, when —x is sub- stituted for x and — y for y, the equation reduces to its original form, the curve is symmetrical with respect to the origin. It is evident that this occurs in an algebraic equation if each term is of even degree, or if each term is of odd degree, in x and y. In applying this test a constant terin is considered as of even degree. It also follows that if a curve is symmetrical to both coordi- nate axes it is symmetrical with respect to the origin. EXERCISES \ State for which of the following equations the curves are symmetrical with respect to the z-axis, the y-axis, and the origin. \ 1. Zx + y + 6 - 0. 7. x* + y = 6. 2. x* + y* = 25. 8. x* + 2xy + y* = 9. 3. 3s* - \y % m 12. 9. y* = (* + 1)(* - 2). 4. x* + y* + 2x = 16. 10. xV + 4x 4 = 16. 5. y» = 4x. 11. Z* + 4z + 2y + 3 - 0. 6. x*y* = 16. 12. x* - x = y. 47. Extent. — Under this heading we endeavor to find how the curve lies with reference to the coordinate axes by finding, first, for what values of either variable there are no points on the curve; and, second, for what values of either variable the curve extends to infinity. To do this the equation is solved for each variable in terms Digitized by Google 52 ANALYTIC GEOMETRY [§47 of the other. First, if a radical of even index involves a variable, certain values of that variable may give imaginary- values for the other variable, in which case there are no points on the curve. If no radicals of even index are involved, there will be at least, one real value of either variable for a real value of the other. In which case there are points on the curve for every value of either variable. Second, if the solution for either variable gives rise to a fraction having the other vari- able in the denominator, then certain finite values of the second variable may make the first in- finite. If no such fraction occurs both variables may become in- finite at the same time. Example 1. — Investigate 9s* + 4y* =« 36 as to extent. Solving for x, x = + js /9 — y* . Solving for y, y = ± \ y/A - x*. Therefore x is imaginary when 9 — y*<0, that is, when y<— 3, or when y > -f 3. And y is imaginary when 4 — x 2 <0, that is, when x< — 2, or when x> +2. The curve is then confined to the portion of the plane in which the abscissas do not exceed 2 in numerical value, and the ordinates do not exceed 3 in numerical value. Example 2. — Discuss the equation x* — ax — y — and plot the curve. Discussion. Intercepts. — Let x = then y = 0. Let y — then x* — ax =» 0. Solving this for x, x — or ± \/a . Hence the ^-intercept is 0, and the x-intercepts are and ± \/o • Symmetry.-- -Since all terms are of odd degree in x and y, the curve is symmetrical with respect to the origin. Extent. — Solving for y % y — .«• — ax. Digitized by Google §48] LOCI AND EQUATIONS 53 The letter a represents an arbitrary constant and may have any value assigned to it. But, in assigning a value, do not choose one that would cause a term to disappear. For the purposes of this discussion it is given the value 4. Since no even root is involved, either variable has a real value for any value of the other. Since large positive values of x make x 3 — ax large and positive, for such values of x, y increases as x increases. Likewise, for numerically large negative values of x, y decreases as x decreases. Plotting. — Tabulating coordinates for positive values of x, the curve can be located in the first and fourths quadrants and, by symmetry, in the second and third quadrants, and is as shown in Fig. 40. The arbitrary value assigned to a is 4, and the unit on the y-axis is one-fifth of that on the x-axis. X i l U 2 3 4 y -ii -3 -2| 15 48 EXERCISES Discuss each of the following equations and plot their curves. 1. ** + y* = 64. 10. xy + 12 = 0. 2. x* - y* - 64. 11. y = X s - 9x. 3. 4x» + 9y» - 36. 12. y(x* + 1) - 8 - 0. 4. 4x* - 9y* - 36. 13. 9x* - y*. 5. y* = &x. 14. y* - (x - 2)(* + 1)(* + 3) 6. x* - Sy. 15. y* = (x - l)*(x - 2). 7. x 2 = Sy - 6. 16. y{x - 1) - 1. 8. re 2 + y* - 4* - 20 = 0. 17. y* = ax 3 -f x 1 . 9. xy = 15. 18. x(x - 2a)* - ay* = 0. 48. Composite loci. — Any function of the two variables x and y may be denoted by f(x, y). Then f(x, y) = is a compact way of writing any equation in these two variables. Theorem. — If the expression f(x, y) can be factored into variable factors, the locus of f(x, y) = consists of as many distinct curves as there are variable factors off(x f y). Proof. — Suppose /(x, y) can be factored into/i(x, y), /*(#, y), M*> V), ' ' ' • Digitized by Google 54 ANALYTIC GEOMETRY IH9 Then AC*, »)/,(*, y)/,(«, y) • • - =0. Now any values of x and y that will make any one of these factors equal zero will satisfy the original equation. Hence all points on the separate loci of M*,v) - 0,/i(«, y) - 0,/,(z, y) - 0, • • • will also be points on the locus of /(*> y) - o. Much time is often saved when /(#, y) can be factored, by plotting each of the equations fi(x, y) = 0, /*(*, y) - 0, /,(», y) - 0, • • • separately. +~x Fig. 41. Example, — Plot the locus of the equation a* + xy* - 2x* - 2y l - 16x + 32 = 0. The factors of x* + xy* - 2s 1 - 2y* — 16s + 32 are (* - 2)(x* +y» - 16). Equating each factor to zero, x — 2 and a* + y* = 16. The first is a straight line and the second a circle as shown in Fig. 41 EXERCISES Find a single equation whose locus is the combination of the loci of the separate equations in each of the following and plot. 1. xy - 6 = 0, xy + 6 = 0. 2. x - 2y + 3 = 0, x - 2y - 3 = 0. 3. x = 0, x = 3, x = 5. 4. x - y y x * + y* - 16. 5. jc* -+- y* = 4, sy = 6. Plot the locus of each of the following by first factoring /(re, y), and then plotting each factor equated to zero. 6. a5*y* - 16. 8. x 1 + 2xy + y* - 4 - 0. 7. x* - y* - 0. 9. x* - 6s* + 11* - 6 * 0. 10. x* + x*y - 4x - 4y + xy* + y» = 0. 11. Plot the locus of (x* — x — 6)(y* + 2y — 8) =0, and show that the lines enclose a rectangle. 49. Intersection of two curves. — The curves of two equa- tions are, in general, distinct, and may or may not intersect. Digitized by Google §50] LOCI AND EQUATIONS 55 It follows from the definition of the locus of an equation that, if a pair of values satisfy both equations, they are the coordi- nates of a point of intersection. And, conversely, if the curves intersect, the coordinates of a point of intersection must satisfy both equations. In order then to find the coordinates of the points of inter- section, it is only necessary to solve the equations simultane- ously. Or, in order to find values of x and y that satisfy the equations simultaneously, the equations may be plotted and the codrdinates of the points of intersection determined from the figure. This is useful when the equations are such as cannot be solved simultaneously. EXERCISES Find the points of. intersection of the curves of the following pairs of equations by solving the equations. Check the results by plotting. 1. x 1 + y* - 16, x + y » 0. 2. x* + y* - 16, x* - y* = 9. 3. x* - 4y l + 7 = 0, 2x + 3y - 12 = 0. 4. x* + y l - 25, 9s* + 49y l = 441. 5. x* + y l - 25, 27y* - 16rr*. 6. Find the distance between the points of intersection of x* + y 2 * 12, and y* — 4x. 7. Solve the following equations by plotting to find the codrdinates of the points of intersection: x 1 + y = 7, x + y % =* 11. EQUATIONS OF LOCI 60. So far in the present chapter the problem considered has been the finding of the locus when the equation was given. Here the second fundamental question is taken up, that of finding the equation of a locus when the locus is known. That is, the algebraic statement is to be found when the geometric figure or description is known. Definition. — The equation of a locus is an equation such that (1) the codrdinates of every point on the locus satisfy the equation, and (2) every pair of values which satisfy the equation are the codrdinates of a point on the locus. Digitized by Google 56 ANALYTIC GEOMETRY [§51 61. Derivation of the equation of a locus. — The process of deriving the equation of a locus depends largely upon the ingenuity of the individual. The following suggestions, how- ever, will be helpful, but it is not intended that it is necessary always to take these steps in order. (1) From the description of the locus sketch a figure involving ail the data. (2) Draw a pair of codrdinaie axes and select P(x, y) any point on the locus. Frequently the coordinate axes are determined by the data; but if they are not, they should be located so as to make the equation as simple as possible. (3) Write an equation between geometric magnitudes, using the conditions of the problem. (4) Express the geometric magnitudes of (his equation in terms of the codrdinates of P and the given constants, and simplify the resulting equation. The final equation will, in general, contain the variables x and y, and all the constants involved. (5) Show that any point whose codrdinates satisfy the equation, is on the locus, and thus show that the second requirement of the definition is fulfilled. A discussion of the equation will often give further facts concerning the locus. Example 1. — The locus of a point is a straight line passing through Pi (—2, 3) and having an inclination of 60°. Find its equation. Solution. — (1) Here the coordinate axes are determined by the data. In Fig. 42, OX and OF are the axes and PiP the locus. (2) P(x t y) is any point on the locus. (3) Slope PiP = tan 60°. »»x Fig. 42. (4) Rope PS -j^-? by [6]. ■ F-3 ' 'x +2 Simplifying, Zx - y/Zy + 6 + Z\/Z VI. Digitized by Google §51] LOCI AND EQUATIONS 57 (5) Any point P(x, y) whose coordinates satisfy the equation 3x - y/Zy + 6 + 3 y/Z - 0, must also satisfy the preceding equation since this equation can be reduced to that form by reversing the steps. But the equation 2,-3 V3 +~X x-f2 simply says that the slope of a straight line through (x, y) and (—2, 3) is equal to y/Z> and hence its inclination is 60°. Therefore the equation of the locus is Zx - y/Zy + 6 + 3\/S = 0. Example 2. — Find the equation of the locus of a point that moves at a dis- tance 8 from the point (3, —5) and re- mains in the plane of the coordinate axes. Solution. — (1) In Fig. 43, the codrdi- p , x y ^ nate axes are drawn and the data located. (2) P(x, y) is any point on the locus. (3) By the conditions of the problem, PC = 8. But PC = y/{x - 3)* + (y + 5)» by [3]. (4) .*. V(x - 3)» + (y + 5) 2 - 8. Squaring, z J - 6x + 9 + y % + lOy + 25 - 64. Simplifying, x 1 + y* - 6x -f 10y - 30 - 0. This is the equation that is satisfied by the coordinates of any point on the locus. The proof of the converse is left to the student. EXERCISES Give orally the equations of the loci described in exercises 1 — 10. 1. A point moves parallel to the y-axis and 4 units to the right. Parallel to the y-axis and 6 units to the left. 2. A point moves parallel to the z-axis and 7 units above. Parallel to the x-axis and 3 units below. ' 3. A point moves parallel to the z-axis and 3 units above the point (3, 6). Parallel to the a>axis and through the point (—6, 4). Parallel to the a>-axis and through the point (0, —7). 4. A point moves parallel to the line y - 4 and 6 units above it. 5. A point moves parallel to the line x = — 3 and 8 units to the right of it. Digitized by Google 58 ANALYTIC GEOMETRY [§51 6. A point moves so as to bisect the angle the y-axis makes with the x-axis. 7. A point moves so as to bisect the angle the x-axis makes with the y-axis. 8. A point moves so as to keep 6 units from the origin. 9. A point moves so as to keep 8 units from the point (2, —1). 10. A point moves so as to keep equidistant from the lines y ■= 8 and j/ =» —2. 11. Find the equation of the locus of a point that is equidistant from the points (5, 4) and (—6, —2). 12. Find the equation of the locus of a point that moves at a distance 10 from the point (-6, -8). 13. Find the equation of the circle having its center at the point (3, 4), and passing through the point (7, 7). 14. Find the equation of the circle having the extremities of a diameter at the points (-4, -6) and (2, 2). 15. Find the equation of the perpendicular bisector of the line joining the points (—4, -8) and (5, 2). 16. Find the equations of the perpendicular bisectors of the sides of the triangle whose vertices are the points (0, 0), (8, 6), and (—4, 10). 17. Find the equation of the locus of a point that moves so as to keep four times as far from the x-axis as from the y-axis. Plot. 18. Find the equation of the locus of a point that moves so as to keep three times as far from the point (2, 3) as from the point (—6, 2). 19. A point moves so that its ordinate always exceeds ) of its abscissa by 8. Find the equation of its locus and plot. 20. A point moves so that the sum of its distances from the points (3, 0) and ( — 3, 0) is 8. Find the equation of its locus and plot. 21. A point moves so that the difference of its distances from the points (3, 0) and (—3, 0) is 4. Find the equation of its locus and plot. 22. A point moves so that the difference of the squares of its distances from the points (—3, —1) and (—2, —4) is 5. Find the equation of the locus and plot. 23. A point moves so that the slope of the line joining it to the point (—2, 3) equals twice the slope of the line joining it to the point (4, —2). Find the equation of the locus. Digitized by Google CHAPTER IV THE STRAIGHT LINE AND THE GENERAL EQUATION OF THE FIRST DEGREE 62. Conditions determining a straight line. — In plane geometry it is found that two independent conditions deter- mine a straight line. Just so in analytic geometry any two conditions that fix the line will determine its equation. Since the same straight line can be determined in a number of different ways, it may be expected that there will be several forms of the equation for the same straight line. Some of the conditions that de- termine a straight line are the following: (1) A point on the line and the direction of the line. (2) Two points on the line. (3) The length and direction of the perpendicular from the origin to the line. Each set of these conditions gives rise to a standard form of the equation of a straight line. 63. Point slope form of equation of the straight line. — Suppose the straight line I, Fig. 44, passes through the point P\(xi, yi), and that its direction is given by its slope m = tan a. If P(x 9 y) is any point on I, then the slope of PPi must be constant and equal to m. By [6], the slope m of PPi is m = V ~ Vl - X — X\ Clearing this equation of fractions, [16] y - yi = m(x - Xi). 59 Fio. 44. Digitized by Google 60 ANALYTIC GEOMETRY [§54 *»x This is the point slope form of the equation of a straight line. Since P(x, y) is any point on I it follows that evejy point on I satisfies [16]. In order to prove that every point which satisfies [16] is on line I, let Pstes, yz) 9 Fig. 45, be such a point, then yz - yi = rn{x z - Xi). Dividing both sides of this equation by X* — Xi, y*- y\ = m. Xz — X\ This shows that the slope of the line PiP 8 = m. Therefore PiP 3 and I are parallel. Since PiPa and I pass through the same point Pi, the line P1P3 and I coincide. There- fore Pz lies on I. In the discussion of other forms of the equation of a straight line, the proof that every point whose coordi- nates satisfy the equation of the locus, is on the locus, is so similar to the proof just given that it will be omitted. Nevertheless this fact should not be lost sight of, for it is one of the essential conditions in determining the equation of a locus. 64. Lines parallel to the axes. — In article 53 it is tacitly assumed that the line whose equation is to be found is not parallel to the y-axis. If it is, a equals 90°, m is infinite, and equation [16] is meaningless. If the line is parallel to the #-axis, it must cut the x-axis at some point (a, 0). Every point on this line has its abscissa equal to a, hence the equation of the line is x = a. Similarly every line parallel to the z-axis cuts the y-axis at some point, say (0, 6). Every point on this line has its ordinate equal to b and hence the equation of the line is y = b. Fio. 45. Digitized by Google §66] EQUATION OF THE FIRST DEGREE 61 Example l.—Find the equation of a line through (—2, 3) and with an inclination of 136°. Substituting x\ — — 2, y\ — 3, and m = tan 136° — — 1 in [16], y-3 - <-l)<* + 2), or x + y — 1 - 0. Example 2. — Find the equation of a line through the point (2, 6) and parallel to the line joining the points (—3, 4) and (1, 5). By [6] the slope of the line joining the two points is J. Therefore the slope of the required line is also } Substituting m — }, Xi « 2, and y\ =» 6 in [16], the equation of the required line is y — 6 - \{x — 2), or x - 4y + 22 = 0. EXERCISES Find the equations of the lines determined by the following sets of conditions: 1. Through (2, -3), slope J. 2. Through (-2, -4), inclination 135°. 3. Through (1, 5), inclination 120°. 4. Through ( — 1, 2), parallel to the line joining (7, 6) to (2, 3). 6. Through ( — 1, 2), perpendicular to the line joining (7, 6) to (2, 3). 6. Through (3, 4), parallel to the t/-axis. 7. Through (3, 4), parallel to the s-axis. Fio. 46. 8. Through ( — 1, 2), inclination — tan -1 J. 9. Through (1, —2), inclination = sin" 1 # . 10. Through (3, 2), inclination = cos -1 ^. 11. Find the equation of the tangent line to the curve y — «* — x, at the point whose abscissa is 2, if its slope equals 11. Suggestion. — Find the ordinate of the point whose abscissa is 2 and substitute in [16]. 12. Find the equation of the tangent line to the curve y =■ 2x* — x + 3 at the point whose abscissa is 2, if its slope equals 7. 65. Slope intercept form. — In Fig. 46, let the intercept of the line on the #-axis equal 6 and let the slope of the line equal m. Since the y-intercept has the coordinates (0, b) this problem is a special case of the point slope form. Putting xi = and j/i = 6 in [15], then y — 6 = mx f or [16] y = mx + b. Digitized by Google 62 ANALYTIC GEOMETRY [§56 This is the slope intercept form of the equation of a straight line. 66. Two point form. — Let the two points through which the line passes be P\(x\, y\) and Piixi, y*). Since Pi is a point on the line and m is the slope of P1P2, this form can be derived V\ — Vt from [16] by substituting for m its value — . Equation X\ %% [16] then becomes [17] y - yi - |^r < x " Xl >- Xi — Xj Note that this equation is not valid if X\ — x% = 0. The line is then parallel to the t/-axis and hence its equation is x = X\. This is the two point form of the equation of a straight line. Since the three points P, Pi, and Pi on the straight line through PiP 2 , always form a triangle whose area is zero, the equation of the straight line can be written in the deter- minant form by article 34, as follows: = 0. EXERCISES Find the equations of the lines given by the following sets of conditions: 1. The y-intercept « 3 and the slope = J. 2. The y-intercept = — 2 and the slope = 3. 2 3. The y-intercept = \ and the inclination = sin -1 — =• Vl3 4. Passing through the points (1, 6) and (7, 2). 5. Passing through the points (—2, 1) and (3, —4). 6. Passing through the points ( — 1, —2) and (—4,-3). 7. What is the effect on line [16] if b is changed while m remains unchanged? What is the effect if m is changed while b remains unchanged? 67. Intercept form. — If the straight line cuts both axes, let its ^-intercept, Fig. 47, equal a and its ^-intercept equal 6. Its equation can be derived from [17] by replacing (xi, yd by (a, 0) and (x 2 , y 2 ) by (0, b). X y 1 Xi V\ 1 X2 y* 1 Digitized by Google §57] EQUATION OF THE FIRST DEGREE 63 Equation [17] then becomes y = — (x — a). Multiplying both sides of this equation by j~t and transpos- ing the :c-term to the left hand side, it becomes [18] l+l-> This is the intercept form of the equation of a straight line. Care must be used in employing this form of the equation, since it is not valid if either or. both intercepts are zero. 68. Normal form. — A line is completely determined if the length and direction of the perpendicular to it from the origin are known. Fig. 47. ^^O Fig. 48. Let C, Fig. 48, be the foot of the perpendicular drawn to the line from the origin, and let (p, 6) be the polar coordinates of C Then OC = p and angle XOC = 0. Since the line AB is perpendicular to the line OC, its slope is the negative reciprocal of the slope of OC and equals —cot 0. Digitized by Google 64 ANALYTIC GEOMETRY [§58 Since the line AB passes through the point C, a point on AB is known. The rectangular coordinates of this point are (p cos 0, p sin 6). Hence the equation of AB can be found by substituting in [16], m = — cot 0, X\ = p cos 0, and y x = p sin 0. Making these substitutions, [16] becomes y — p sin 6 = —cot 6 (x — p cos 0). Multiplying both sides of the equation by sin and transpos- ing all terms to the left hand side, gives x cos + y sin — p(sin 2 + cos 2 0) = 0. [19] x cos e + y sin e - p « 0. This is the normal form of the equation of a straight line. If cot = » , the line is parallel to the #-axis and its equa- tion is x = p. But even in this case the normal form is valid, for if cot = » , = 0° and the normal form would read x cos 0° + y sin 0° — p = 0. Since cos 0° = 1 and sin 0° = this equation is equivalent to x = p. Example. — Find the normal form of the equation of a straight line if $ » 30° and p » 6. Substituting these in [19], x cos 30° + y sin 30° - 6 - 0. Since the polar coordinates of C can be written either (6, 30°) or (—6, 210°), the normal form of the equation of this line could also be written x cos 210° + y sin 210° +6-0. That these equations are equivalent can readily be seen if the trigonometric functions are replaced by their numerical values. EXERCISES Find the equations of the following lines having given: 1. a - 3, b - -2. 6. 6 - 60°, p « -3. 2. a - -1, b - 6. 7. - 135°, p = 2. 8. o - i 6 - |. 8. - 210°, p - 1. 4. a - -i 6 - -i 9. - 330°, p - - 4. 6. e - 60°, p - 3. 10. - 150°, p - - 2. 11. What is the effect on line [19] if p is changed while remains un- changed? What is the effect if is changed while p remains unchanged? Digitized by Google §59] EQUATION OF THE FIRST DEGREE 65 69. Linear equation. Theorem — Every equation of a straight line is of the first degree in one or two variables. Conversely. Every equation of the first degree in one or two variables is the equation of a straight line. Proof. — Every straight line intersects, does not intersect, or coincides with the y-axis. In the first case by means of article 55 its equation can be put in the form y = mx + b, in the second case its equation is x = a, and in the third case x = 0. Each of these equa- tions is of the first degree in x and y. Proof of converse. — Consider the most general equation of the first degree in two variables. This is [20] Ax + By + C - 0. Assume that B^O and solve this equation for y, it becomes Ax C * - - "F - F Comparing this equation with the form y = mx + b shows at once that it is the equation of a straight line whose slope m = — -^, and whose y-intercept 6 = — -g- Q If B = 0, [20] becomes Ax + C = 0, or x = - j This is the equation of a straight line parallel to the y-axis. Hence every equation of the first degree is the equation of a straight line. Ax + By + C = 0, the most general equation of the first degree in two variables, is called the general equation of a straight line. 60. Plotting linear equations. — Since every equation of the first degree represents a straight line, it is sufficient in plotting the graph of such an equation to find two points which satisfy the equation and then join these points by a straight line. Usually two such points that can be easily found are the inter- cepts on the x and the y-axes. 5 Digitized by Google 66 ANALYTIC GEOMETRY [§61 61. Comparison of standard forms. — In article 59 it was seen how the general equation could be transformed into the slope intercept form. This method of transforming one form of an equation into another is of great use in analytic geometry, since by comparing the constants in two forms of an equation of a line or curve, much information can be secured. In this particular case the transformation from the general form to the slope intercept form, enables one to read off by inspection the slope and y-intercept of the line. For example, if the equation 3x + 4y = 12, be solved for y it becomes V - -fa; +3, Comparing this equation with y = mx +b f shows that the slope of the line is — f and its ^-intercept is 3. Take the same equation, Zx + 4y =» 12, and divide both sides of the equation by 12, and ^ + g ■■ 1. Comparing this equation with - + ? = 1 shows that the x-intercept is 4 and that the ^-intercept is 3. Here the a>intercept can be as easily found by putting y=Q in the original equation; and the ^-intercept by putting x = 0. 62. Reduction of Ax + By + C = to the normal form. — In general A and B will not be the cosine and sine respectively of the same angle, and hence Ax + By + C = will not be in the normal form. In order to transform it to the normal form multiply both sides of the equation by an arbitrary constant k, whose value is to be computed later. This gives Akx + Bky + Ck = 0. The quantity k is now assumed to be such a number that Akx + Bky + Ck = will be identical with x cos 6 + y sin — p = 0. Comparing coefficients gives Ak = cos 6, Bk = sin 0, Ck = -p. Digitized by VjOOQIC §62] EQUATION OF THE FIRST DEGREE 67 To find the value of k, square both sides of the first two equations and add. This gives A 2 & 2 + B*k* = cos 2 + sin 2 - 1. Solving for A, k = j^—=- If & is replaced by its value, Akx + Bky + Ck = becomes Ax By C ran v 4. J 4. — — a 1 J + VA 2 + B 2 T ±Va 2 + b 2 t ±VA 2 + b 2 This is the general equation of the straight line expressed in the normal form. Either sign can be used with the radical, but of course the same sign must be used throughout the equation. Comparing this equation with the normal form gives [22] C0Se= ±VA^ + B^ Sine= ±VA° + B» P ±VA 2 + B 2# Hence, to transform the equation Ax + By + C = to the normal form, divide both sides of the equation by ±Va* + b*. Example 1. — Change 3x — 4y + 6 = into the normal form. Here A - 3, B - -4, C - 6, and±VA? + £* - ±V9 + 16 - ±5. Dividing the equation through by ±5, it becomes ±5 ±5 ^ ±5 " u * Either sign can be used since the two equations £ x — $ y + $ — 0, and — #£ + £y — £ =0are equivalent. EXERCISES Find the slope and y-intercept of the following equations by express- ing them in the slope intercept form. 1. Zx + 2y - 4 - 0. 3. -5s + 2y - 6 = 0. 2. 2x - 3y 4- 2 = 0. 4. 2s - 2y + 7 - 0. Digitized by Google 68 ANALYTIC GEOMETRY [563 Change the following equations into normal form, and find the distance of the line from the origin. 5. Zx - 4y - 6 - 0. 9. x + 2y - 3 - 0. 6. -3s + 4y + 10 - 0. 10. -Zx + y - 6 - 0. 7. 5x - 12y + 26 = 0. 11. 2x + 3 - 0. 8. -5s - I2y + 39 = 0. 12. Zy - 4 - 0. Find the equations of the lines satisfying the following conditions: 13. Through the point (1, 2) and parallel to Zx - 4y -f 6 - 0. 14. Through the point (2, -3) and parallel to x + 2y - 3 - 0. 15. Through the point (6, 2) and perpendicular to 2x + y — 3 = 0. 16. Through the point ( —3, 1) and perpendicular tos — y + 6=0. 63. Distance from a point to a line. — The aistance from the origin to the line x cos 6 + y sin 6 — p = 0, is the numeri- cal value of p. Hence if d! is the distance from the origin to the line Ax + By + C = where the sign of the radical is chosen so as to make d' positive. N Y' >i(*i.»> ^ In order to find the distance d from the point Pi(xi, yO, Fig. 49, to the line. 4s + By + C = 0, translate the axes to the new origin P*(xi, yi). The equations of trans- lation [12] are x = x' + Xi, Fig. 49. » « »' + Vl* Making these substitutions, the equation Ax + By + C - becomes Ax' + By' + Axi + By x + C = 0, where the new constant term is Axi + By, + XJ. The distance d = PJt is the distance from the new origin to the line. From equation (1) m d . A* + g* + C 1 J ±Va 2 L 2 + B 2 ' where the sign of the radical is chosen to make d positive. Digitized by Google §63] EQUATION OF THE FIRST DEGREE 69 Example 1. — Find the distance from the point (2, 1) to the line 3s - 4y + 6 - 0. Solution. — Translate the axes so that the new origin is the point (2, 1). The equations of translation are s - s' + 2, y - v' + l. The equation of the line 3s — 4y + 6 =0, referred to the new origin is 3(s' + 2) - W + 1) + 6 - 0, or 3s' - ±y' + 8 - 0. Putting this equation into the normal form gives 3s' - V + 8 ±5 Hence the distance from the new origin to the line is d ■* -£, and this is the distance from the point (2, 1) to the line 3s — 4y + 6 — 0. This distance could be found also by substituting directly in [28]. Putting A « 3, B « -4, C « 6, Xi - 2, y x - 1, , 3-2 -4-1 +6 +8 8 ±5 " +5 " 5* Example 2. — Find the distance from the point (3, —2) to the line 5s + 12y - 4 « 0. Putting A - 5, B - 12, C - -4, xi - 3, and y & = -2 in [28], , 5-3 + 12(-2) -4 -13 ±13 " -13 * ' This apparent Inconsistency arses because both signs must first be put down and then the correct sign selected. EXERCISES Find the distances from the points to the lines in the following exercises: 1. Point (2, 3) to line 4s - 3y + 4 - 0. 2. Point (-1, 2) to line Sx + 4y - 6 - 0. 3. Point (1, 3) to line x - y « 0. 4. Point (2, 3) to line x cos 30° + y sin 30° - 3 = 0. 5. Point (3, -1) to line x cos 135° + y sin 135° + 1-0. 6. Point (1, 6) to line y - 1 - 3(s - 4). 7. Find the altitudes of the triangle whose sides have the equations y = 1, 12s + 5y - 27 = 0, and 3s - 4y + 9 = 0. 8. Find the altitudes of the triangle whose vertices have the coordi- nates (4, 2), (-3, 1), (6, -3). Digitized by Google 70 ANALYTIC GEOMETRY [§64 64. The bisectors of an angle.— Let the sides of an angle be formed by the lines ST and SR, Fig. 50, the equations of which are A x x + B x y + C x = and {p(*v) A& + B%y+ Ct = respectively. Let P (x, y) be any point on the bisector SP of the angle formed by these two lines. From plane geometry it is known that the bisector of an angle is the locus of points equi- distant from the sides. Hence PT = PR. Fig. 50. Expressing this fact algebraically gives Axx + Bi y + d AjX + B % y + C 2 ±VA? 5 + 5i 2 ±VA? + B* 2 The four possible combinations of signs in this equation will yield two different equations, [24] AjX + Bjy + Ci = A^x + Bay + C 2 vV + Bx' - Va 2 * + b 2 * ' One of these is the equation of the bisector of the angle RST, while the other is the equation of the bisector of the supplementary angle TSR'. In order to tell which equation belongs to the bisector sought, draw the figure as accurately as possible and observe whether the slope of the required bisector is positive or negative. Since the two bisectors given by [24] are at right angles to each other, one has a positive slope and the other has a negative slope, so that in general it is easy to pick out the required equation. The exceptional case occurs when one bisector is very nearly parallel to the z-axis, and it is diffi- cult to tell the sign of its slope. In this case the numerical value of its slope is small, whereas the numerical value of the slope of the other bisector is large, so that again it is easy to associate the equations with the correct bisectors. Digitized by Google EQUATION OF THE FIRST DEGREE 71 Example. — Find the equation of the bisector of the angle which the line Zi = 3s + 4y — 5 = makes with the line 1% m bx — 12y + 6 - 0. In Fig. 51, let U be the required bisector. By [24] the equations of the two bisec- tors are 3s -+ 4y - 5 - ± bx - 112y + 6 5 13 Clearing of fractions and simplifying gives the two equations 14* + ll2y - 95 - 0, (1) 64x - Sy - 35 - 0. (2) The slope of (1) is small and negative, whereas the slope of (2) is large and positive. Since the slope of 1% is large and positive, its equation is 64x — Sy — 35 = 0. EXERCISES Find the equations of the bisectors of the angle which the first line makes with the second in exercises 1-6. 1. 8* + y - 6 - 0, 7x + 4y - 3 - 0. x - 7y + 6 = 0, bx + by - 8 - 0. 11* - 2y + 12 = 0, 2x + y - 6 = 0. 13* + y - 15 - 0, 22* - Uy - 21 - 0. 12* + Uy - 11 - 0, 9* - 2y + 10 - 0. 9* + 7y - 6 - 0, 11* + Sy - 14 - 0. Find the equations of the bisectors of the angles of the triangle the equations of whose sides are 8* — y + 1 = 0, * + Sy + 1 - 0, and 7* 4- 4y - 43 - 0. 8. Find the equations of the bisectors of the angles of the triangle whose vertices are flfc "¥), (1, «, and (12, -1). 65. Systems of straight lines. — Sometimes the geometrical facts given are not sufficient to determine a straight line uniquely. In such a case not all the constants entering into the equation of the line will be determined. For instance, if the problem is to find the equation of a line that is parallel to 3x + 4y : — 6 = 0, Fig. 52, it is evident that there are an un- 2. 3. 4. 6. 6. 7. Fig. 52. Digitized by Google 72 ANALYTIC GEOMETRY [§66 limited number of lines in the plane which satisfy the condi- tions of the problem. To find the equation of any one of these lines, substitute m = — f, in [16], which becomes y = — lx + 6, or 3x + 4y — 46 = 0. The quantity 6 can have any value whatsoever. If it is given some arbitrary value the equation 3x + 4y — 46 = becomes the equation of some one of the lines that are parallel to Sx + 4y — 6 = 0. All of these parallel lines taken together are said to form a system of lines. Another system of lines consists of all the lines through a given point. If the point has the coordinates (1, 2), the equa- tion of this system of lines is y — 1 = m(x — 2), by [15]. EXERCISES Find the equations of the following systems of lines: 1. All the lines passing through the point (—2, 3). 2. All the lines passing through the origin. 3. All the lines passing through the point (3, 4). 4. All the lines having their x-intercept equal to 3. 6. All the lines haying their y-intercept equal to —4. 6. All the lines at a distance 3 from the origin. 7. All the lines at a distance 7 from the origin. 8. All the lines parallel to the line 2x + y — 3 = 0. 9. All the lines perpendicular to the line x — Sy +' 6 =» 0. 10. All the lines such that the a?-intercept of each is equal to its y-intercept. 66. Applications of systems of straight lines to prob- lems. — Sometimes the facts determining a straight line are not such that its equation can be written down immediately. This happens if the slope m and the distance p of the line from the origin are given. In such a case there are two methods of procedure, one is to compute the constants which oc- cur in some standard form of the equation of a straight line, by drawing the figure and applying plane geometry or trigonome- try. Another method is illustrated in the following example. Digitized by Google §66] EQUATION OF THE FIRST DEGREE 73 Example. — Find the equation of the straight line given m - J, and p = 3. First method. — First write down the equation of the system of lines whose slope is t . This is y « |x + b. Next transform this line into the normal form. Its equation becomes 4x - 3y + 3& ±5 The distance of this line from the origin is 3. 3, and b 36 77, but this is p and p Hence -r-= * ±5 Substitute this value of 6 in y and it becomes y = Jx ± 5, or 4x - 3y ± 15 - 0. There are two lines which satisfy the required conditions, and they are equally Fig. 53. distant from the origin. Second method. — Write down the equation of all lines distant 3 from the origin. This equation is x cos $ + y sin $ — 3 — 0. In order to determine 0, note that the slope of this line is — cot 0. Therefore — cot $ = $ , and $ can be in either the second or the fourth quadrants. If $ is in the second quadrant, then sin $ « f and cos $ « — 1« If 6 is in the fourth quadrant, then sin $ — — i and cos 6 =» |« Substituting these values, the equation be- comes ± \x T $ y — 3 - 0. Multiplying both sides by ±5, gives 4x - Zy ± 15 - 0. Example 2. — Find the equation of a line ^*-£ through the point (1, 3) and making equal intercepts on the axes. First solution, geometric method. — In Fig. 54, Fig. 54. let AB be the line through Pi(l, 3) whose equa- tion is to be found. Since the intercepts are equal, angle BAO » angle OB A = 45°. Hence a = OM + MA - OM + MP X -1+3-4, and this is also the value of 6. x 1/ Therefore the required equation ^4 + 4 = 1, orx + y — 4. Unfortunately by using the geometric method parts of the solution are Digitized by Google 74 ANALYTIC GEOMETRY [§66 liable to be overlooked. This is illustrated very well in this problem, since the line OP\ passing through the origin and the point (1, 3), satisfies all the conditions of the problem and is therefore also a solution. Second solution, algebraic method. — Since the line AB passes through the point (1, 3) its equation is y — 3 ■= m(x — 1). The ^intercept of this line is — — — > and the ^-intercept is 3 — m. Since these are equal, — - — » 3 — m. Solving this equation gives m = 3 or — 1. If to — 3 the equation is y — 3 = Z(x — 1) or y — Sx = 0. If to = — 1 the equation is y — 3 = — (x — 1) or £ + y — 4=0. ThircTsotution, algebraic method. — This differs from the preceding only in that it starts from the intercept form of the equation of a straight line, instead of from the point slope form. Since the intercepts are equal, the intercept form of the equation is - + - = 1. In order to make this line pass through the point (1, 3), 1 3 substitute these coordinates for x and y. This gives — h - = 1. a ■ a Solving,, gives a — 4 and the required equation is j + | — 1, or x + y = 4. The question naturally arises, what happened to the solu- tion y — 3x = 0? This is certainly a solution since the intercepts a = and b = are equal and the line passes through the point (1,3). This question can be answered by noting as stated in article 57 that the intercept form is not valid when either or both intercepts are 0. Hence the solution y — 3x = cannot be secured from the intercept form. Whenever this form of the equation of a straight line is used the question as to whether either or both intercepts are zero must be answered independent of the equation. EXERCISES Find the equations of the lines determined by the following conditions: 1. The slope of the line equals — } and it is distant 1£ units from the origin. 2. The line makes equal intercepts on the axes and passes through the point (4, 2). Digitized by Google §67] EQUATION OF THE FIRST DEGREE 75 3. The line passes through the point ( — 7, 4) and is tangent to a circle whose center is the origin and radius equal to 1. 4. The line passes through the point (4, 2) and is tangent to a circle whose center is the origin and radius equal to 2. 6. The slope of the line is 2 and its z-intercept equals 3. 6. The slope of the line is —2 and the sum of its intercepts is 9. 7. The slope of the line is — } and the sum of its intercepts is 5.. 3. The line makes intercepts which are equal numerically but opposite in sign, and passes through the point (6, 3). 9. The line passes through the point (1, 3) and the sum of its intercepts equals 8. 10. The line passes through the point (3, 1) and the portion in- cluded between the axes is bisected by this point. 11. The line passes through the point (3, y/Z) and the perpendicular from the origin on the line has an inclination of 60°. 12. The line is perpendicular to the line 4z + 3y — 6 — and distant 2 units from the origin. 13. The line is distant 3 units from the origin and its y-intercept equals 5. 14. The line is distant 2 units from the origin and the product of its' intercepts is y. 16. The line passes through the point (1, 2) and makes with the axes a triangle in the first quadrant whose area equals 4. 16. The line passes through the point (1, 2) and makes with the axes a triangle in the second or fourth quadrants whose area equals 4. 67. Loci through the intersection of two loci. — Theorem. Iff(*> V) = and g(x, y) = are the equations of any two loci and Y k is any constant not zero, then o(*.v)^o^ f(*> y) + tyix, y) = is the equa- tion of a curve which passes through all the points of inter- section of f(x, y) = and 9(&> y) — 0, but does not intersect these curves in any other point Proof. — Let Pi(xi, j/i), Fig. 55, be any point of intersection of f(x, y) = and g(x, y) = 0. Since Pi lies on both these curves its coordinates must satisfy each equation, therefore f(x u yd = and g(x u y x ) =0. /r*.v)-o *-* Digitized by Google 76 ANALYTIC GEOMETRY [§67 Substituting the coordinates of Pi in /(a, y) + kg(x, y) = 0, /(*it Vi) + kg(x lt y x ) - + fc = 0. Therefore Pi lies also on the curve f(x, y) + kg(x, y) = 0. But Pi was any point of intersection of f(x, y) = and Q&i y) a 0, therefore every point of intersection of these curves lies on f(x, y) + kg(x, y) = 0. Furthermore the curve /(x, y) + kg(x, y) = cannot meet either f(x, y) = or g(x, y) = in any other point. For if it did, suppose it meets /(x, y) = at Pj(x2, 2/2), and that P2 is not on g(x, y) = 0. Then /(x 2 , j/2) = 0, but g(x 2 , #2) = a where Substituting the coordinates of P 2 in /(a;, y) + A#(a;, y) = 0, /(*a, y«) + kg(x t , y 2 ) = + ka 5* 0. In like manner it can be shown that f(x y y) + fcgrfo y) = will meet g(x, y) = only at the points of intersection of /(a, y) = and y(x, y) = 0. If /(x, y) = is the straight line Ax + By + C = 0, and g( x > y) — is the straight line it's + B'y + C" = 0, then /Or, y) + fcg(x, y) = 0, or Ax + By + C + k(A'x + B'y + C") = is the equation of a straight line through the point of in- tersection of the straight lines, Ax + By + C = and A'x + B'y + C" - 0. Example 1. — Find the equation of the straight line which passes through the point (4, 3) and through the intersection of the two lines 2x + 3y - 5 - and 3s - 4y + 1 - 0. It has just been shown that the equation of any line passing through the intersection of these two lines is of the form 2x + Zy - 5 + k(Zx - 4y + 1) - 0. Since this line passes through the point (4, 3), its equation is satisfied when x — 4 and y = 3. This gives 12 + fc(l) - 0. Therefore A; «■ —12, and the required equation is 2x + 3y - 5 - 12(3* - 4y + 1) - 0, or 2s - Sy + 1=0. Digitized by Google §68] EQUATION OF THE FIRST DEGREE 77 EXERCISES Find the equations of the lines satisfying the following conditions. 1. Passing through the point of intersection of 2x + Zy — 3 — and Zx — y — 1 — 0, and through the point (1, 1). 2. Passing through the point of intersection of 5x — 4y — 2 — and 2x + 4y — 15 ■» 0, and through the point (2, 3). 3. Passing through the point of intersection of Zx + 2y — 6 — and 3 + y ™ 3, and perpendicular to 2a; + !/ — 1 ■*(). 4. Passing through the point of intersection of x — 6y — 3 and 2a: — y *■ 2, and perpendicular to 3 — 2y + 1 «■ 0. 6. Passing through the intersection of y - 6 + x and 3y — 4 — 2x, and parallel to x + Zy — 4 — 0. 68. Plotting by factoring. — Since it is easy to plot a straight line, the theorem of article 48 gives a simple method of plotting equations which can be factored into linear factors. . Example. — Plot the equation 2x* + 2x + 7y - xy + Zy 1 + 4. . First transpose all terms to the left hand side of the equation 2x* + 2x + 7y - xy - 3y* - 4 - 0. In order to find out if this equation can be factored, regard it as a quadratic in x or y, and solve for that variable. For the sake of convenience the variable chosen this time will be x. Collecting like powers of x, 2s* + (2 - y)x - 4 + 7y - 3y* - 0. Solving for x, by means of the formula, Art. 4, where a = 2, 6 ■» 2 — y, and <j a -4 + 7y - 3y*, ± V25y« - 60y + 36 _ -2+y ± (5y - 6) , *-x Fiq. 56. '- 2 + y 4 4 Hence x — -^ — or — y + 1, and the left hand side can be factored in to2(x-^li)( a;+y -l). The equation now becomes (2x — Zy + 4)(x + y — 1) « 0. Therefore the graph of 2s* + 2x + 7y = ^ry + 3y* + 4 consists of the two straight lines 2x — Zy + 4 = and x + y — 1 - 0. When an equation of the second degree in each of two variables, is solved for one variable in terms of the other, an Digitized by Google 78 ANALYTIC GEOMETRY [§69 expression is obtained under a radical sign. If this expression is a perfect square, the graph of the equation consists of two straight lines. Thus, in the problem just solved, 25y* — 60y + 36 is a perfect square. If it had not been a perfect square 2x* + 2x + 7y — xy + 3y* + 4 could not have been plotted by this method. Example 2. — Plot the curve x*y =* y». Transposing all terms to the left hand side, x*y — y» = 0. • Factoring the left hand side, y{x — y)(x + y) — 0. The graph consists of the line y = which is the x-axis, the line x — y = o and the line x + y » 0. EXERCISES 1. Find the equation of the triangle whose sides are x — y f y — 0, and x + y = 1. 2. Find the equation of the square whose bounding lines are x *= 1, i»2,y»l, and y = 2. Plot the following curves by first factoring: 3. rr* - 2y* - *y + Sy - 1 - 0. 4. 2y* = xy + x*. 6. x 2 + 2x + 1 - 4y*. 6. 2x* + xy + 4x + y + 2 - y*. 69. Straight line in polar coordinates. — In general the equations of straight lines in polar coordinates are not as simple as those in rectangular co- ordinates. The simplest case is the one in which the known quantities are the polar coordinates of the foot of the perpendicular from the origin to the line. This is the same data as was given for the normal form of the equation of a straight line. If the end of the perpendicular to the line from the origin has the coordinates (pi, 0i), Fig. 57, let P(p, 6) be any point on the line. Then ' gp cos POPi = nff> or cos (e - *i) = ~ Hence OP ' ~ l WD vv V1/ P - 90 = px. (1) Digitized by Google §70] EQUATION OF THE FIRST DEGREE 79 This equation can also be obtained from the normal form of the equation of a straight line by replacing p by p l9 by 0i, x by p cos 0, and y by p sin 0. In the special case where the line is perpendicular to the polar axis, B x = 0, and the polar form of the equation of the straight line takes the form p cos = pi. If the straight line is parallel to the polar axis, 0i = 90° and the equation of the straight line becomes p sin = p x . Example. — Find the polar form of the equation of a straight line if the coordinates of the foot of the perpendicular drawn to it from the origin are (3, 60°). Substituting in equation (1), gives p cos (0 — 60°) = 3. EXERCISES Write the equations of the following straight lines in polar coordinates, if the coordinates of the end of the perpendicular from the origin to the line are: 1. (3, 45°). 3. (7, 90°). 6. (-4, 135°). 2. (-2, 60°). 4. (4, 180°). 6. (3, 315°). Change from rectangular to polar coordinates. 7. x + y - 1 - . 1 . xVS + y - 4. 8. x - 3. 11. x - yy/3 +6=0. 9. y = -7. 12. y - 2x - 0. Change from polar to rectangular coordinates. 18. p = 3 sec 0. 18. p - -. — w—. — - • 4co8 0-6 8in0 14. p = 4 esc 6. 19. 5 sin 6 — 3. 15. tan $ - 6. 20. 13 cos 6 = -5. 1ft 2 V2 cos + sin p sin (6 - 45°) 4IT 3(cos 6 — sin 0) fto 3 17 » P * — o/i **• cos 20 K cos (0 + 60°) 70. Applications of the straight line. — Whenever two vari- ables are related so that one varies directly as the other, or so that a change in one varies directly as the corresponding change in the other, the relation between the variables is linear, and the graph showing the relation between the variables is a straight line. Since many of the relations in physics, mechanics, and Digitized by Google 80 ANALYTIC GEOMETRY [§70 engineering are of this nature, the straight line has a wide field of application. Oftentimes the curves representing the relation between physical quantities are within certain limits so nearly straight lines that the more complicated equation is replaced, on account of its simplicity, by the linear relation. A few specific instances are the following. (1) The increase in velocity of a body falling under the action of gravity is proportional to the time. This is expressed by the relation v — t> = k(t — Jo), where vo is the velocity of the body at the time t and v is the velocity of the body at any time t. If v and v are expressed in feet per second, and t and to are expressed in seconds, then k, the proportionality factor, is the familiar constant g. This relation is often expressed v = kt + v 0f where v is the velocity when t = U = 0. (2) Hooke's Law. — The extension of an elastic string varies directly as the tension. This is expressed by the relation I = kt + Jo, where I is the length of the string under the tension t y and U is the length of the string when t = 0. (3) The expansion of a bar due to heat, is very nearly pro- portional to its increase in temperature. This is expressed by the relation I — Z = k(t — t ), where h is the length of the bar at some temperature t Q and I is its 1 length at any temperature t (4) The weight of a column of mercury in a barometer varies directly as its height. This is expressed by the relation 10 = kh, where the weight w is taken as zero when h = 0. In all four cases the graph representing the linear relation between the variables is a straight line. Digitized by Google §70] EQUATION OF THE FIRST DEGREE 81 For further applications of the straight line see Chapter X on empirical equations. GENERAL EXERCISES 1. Translate the following algebraic statements into words and draw their loci: (1) V - 4s, (2) y - 4* - 4, (3) x - 3y + 2, (4) x - 5y - 2. 2. Find the equation of the line (1) through the point (4, —3) and parallel to 2x — 3y — 4; (2) through the point (5, 7) and perpendicular to 2x + 7y - 14. 3. Find the equation of the line (1) through the point (—2, —5) and parallel to x — 7y — 3; (2) through the point (h, k) and parallel to the line y = mx + 6. 4. Find the length of the following perpendiculars: (1) From (3, 2) to 4x - 3y - 7 - 0. (2) From (0, -3) to 5x - y - 6 - 0. (3) From (2, 3) to Ox - % - 10 - 0. 6. Find the lengths of the three altitudes of the triangle whose- vertices are (4, 5), (—2, 2), and (3, —4). 6. Find the distances from the line 2x + 3y — 12 = to each of the points (4, 4), (2, -3), (0, 0), (-3, 5), and (-2, 8). 7. Given 4x -f ky — 5 =0; determine the value of k for which the line will (1) pass through the point (—4, 3), and (2) be parallel to 3* - 2y + 7 - 0. 8. Find the equations of the lines through the intersection of the lines 2s -|- y — 16 = and x — y -f 2 = and also (1) passing through the point (2, 7), (2) parallel to the line 7x - 2y + 6 = 0, (3) perpendicular to the line 3a; — 4y + 2 = 0, (4) having the slope ~f. 9. Given a triangle having as vertices the points (6, 2), (—3, 5), and ( — 1, —3); find the equations of the perpendicular bisectors of the three sides, and the coordinates of their point of intersection. 10. Show that 15a:* — 14xy — 8y* = is the equation of two straight lines intersecting at the origin. 11. Prove that, if A, B, and C are real numbers, Ax* + Bxy -f Cy* - represents two straight lines passing through the origin, and that these lines are real and distinct, real and coincident, or imaginary according as B* — 44 C is positive, zero, or negative. 6 Digitized by Google 82 ANALYTIC GEOMETRY [§70 12. The perpendicular drawn from the origin to a line makes an angle of 60° with the s-axis and its length is 2, find the equation of the line. 13. Write the equations of the following lines: (1) Passing through the point (3, 5) and having an inclination of 45°. (2) Passing through ( — 1, —3) and having a slope of 2. (3) Passing through (—2, 8) and having an inclination of 120°. 14. Show that the following lines form a parallelogram: (a) 2s + 3y - 10, (6) 2s + Zy - 20, (c) s - 2y - 5, (d) 2x - 4y - 17. 15. Write the equation of the line passing through the intersection of x — 3y + 8 = and 3s -f 2y + 2 = and making an angle whose tangent is 2 with the s-axis. 16. Find the coordinates of the point in which the perpendicular to the line 2x — y — 1 = and passing through ( — 2, 3) intersects that line. 17. What does the equation 3s — 2y + 4 =0 become when the coordinate axes are turned through an angle of 45°? Plot the locus of the equation in both cases. 18. Plot each of the following lines, translate the axes so that the new origin shall be at the point indicated and replot from the new equation. (1) y = 3s + 4, (2, 3). (3) y - mx + 6, (c, d). (2) 2y - 3s - 2 - 0, (-2, 3). (4) y - 4s + 5 - 0, (}, -2). 19. The three vertices of a triangle are (8, 2), (4, 8), and (— 2,-6). Find the equations of the lines each of which bisects two sides of the triangle. 20. Given two straight lines each having an inclination of 45° and having intercepts on the y-axis of 6 and —8 respectively; find the equa- tion of the straight line that is equidistant from the two lines. 21. Find the equation of a straight line such that the perpendicular from the origin to it equals 8 and makes an angle of 45° with the s-axis. 22. Find the equation of the straight line which passes through the intersection of the lines s — 2y — 4 = and s + 3y — 8 = and is parallel to the line 3s + 4y = 4. 23. Find the equation of the line through the point (2, 3) making an angle tan" 1 } with the line 2s — 4y + 7 = 0. 24. Find the equation of the line through the point ( — 1, 2) making an angle sin" 1 J with the line s + 3y — 4 — 0. 25. Find the equation of the line through the point (6, 4) making an angle cos" 1 (—J) with the line 2s — y + 6 — 0. 26. Find the equation! of the two lines through the point ( — 1, —3), which form an equilateral triangle with the line s + y » 2. Digitized by Google §70] EQUATION OF THE FIRST DEGREE 83 27. Find the equation of the line through the point (0, 6) which together with the {/-axis as the other equal side forms an isosceles triangle with the line 2x — y + 4 = 0. 28. Find the equation of the line through the point (0, 6) which together with the 2/-axis for the other leg forms an isosceles triangle with the line 2x + y - 4 = 0. 29. The equations of the two equal sides of an isosceles triangle are x — 2y + 6 — and 2x — y — 2 = 0. Find the equation of the third side if it passes through the point (9, 4). 30. Find the slope of the line 2x + 3y — 4 = after the axes are rotated through 30°. 31. Find the slope of the line x — Zy + 6 = after the axes are rotated through the angle 0, where cos 6 — — J and 0is in the second quadrant. 32. Find the equations of the two lines through the point ( — 1, 3) which trisect that part of the line 2x + y — 6 — which is intercepted between the axes. 33. An equilateral triangle lies wholly in the first quadrant. If one side has its extremities at (1, 6) and (6, 1), what are the equations of the other two sides? 34. An isosceles right triangle is constructed with its hypotenuse along the line 2x + y — 6 = 0. If its vertex is the point (3, 4), find the equations of its sides. 36. A circle is inscribed in the triangle the equations of whose sides are x + 2y - 16 = 0, 2x - y + 3 = 0, and 2x + y - 7 - 0. Find its radius and the coordinates of its center. 36. The base of an isosceles triangle is the line joining the points (1, 5) and (4, 6), its vertex is on the line x + y — 7 = 0. Find the codrdi- nates of its vertex. 37. Find the locus of a point which moves so as to be always equi- distant from the points (3, 5) and ( — 1, 7). 38. Find the equation of the locus of a point which moves so that its distance from the line 7x + 4y — 6 — is twice its distance from the line x - Sy + 3 = 0. 39. Find the equation of the locus of a point which moves so that the difference of the squares of its distances from the points (—2, 3) and (1, 6) shall be constant and equal to 2. 40. Find the equations of two lines through the point (1,1) such that the perpendiculars let fall from the .point (1, 3) on them are each of length f . 41. Prove that the feet of the perpendiculars let fall from the point (3, 1) on the sides of the triangle x = 0, y =» 0, and 2x + y — 4 = lie in a straight line. Digitized by Google 84 ANALYTIC GEOMETRY [§70 42. Find the equations of the straight lines through the point (3, 6) and intersecting the line x+y — 2=Q&t& distance 5 from this point. 48. Prove that the perpendicular bisectors of the sides of a triangle meet in a point. 44. Find the equation of the locus of a point that is always twice as far from the origin as from the s-axis. 46. The coordinates of two points are (3, 5) and (4, 4). Find the equation of a straight line which bisects the line segment connecting these points and makes an angle of 45° with the s-axis. 46. A straight line inclined to the s-axis at an angle of 150° has an s-intercept equal to 8. Find the equation of a straight line passing through the origin and bisecting that portion of the line included between the axes. 47. Find the equations of the four sides of a square two of whose opposite vertices are (2, 3) and (3, 4). 48. A straight line moves so as to keep the sum of the reciprocals of its intercepts on the axes a constant. Show that the moving line passes through a fixed point. 49. Find the equation of the straight line passing through the point (2, 6) and making an angle of 30° with the line s — 2y » 1. 60. Find the equation of a straight line passing through the point (c, 0) and making an angle of 45° with the line bx — ay = ab. 61. The equation of a straight line is 3s + by = 15; find the equa- * tion of the same line referred to parallel axes whose origin is at (3, 2). 62. Find the equations of the straight lines bisecting the angles formed by the lines 12s + 5y = 8 and 3s — 4# = 3. 68. Show that an angle of 45° is formed by the lines represented by the equation x* — xy — %* + 2x — y + 1 =0. 64. Given the equation Ax + By + C — 0. Find the relation be- tween A, B and C, (1) so that the s- and y-intercepts shall be equal; (2) so that the inclination of the line shall be 45°; (3) so that the line shall pass through the point (1, 2). 66. Determine the angle that the first line of each of the following pairs makes with the second: (1) x + 2y = 5, 3s - 4|/ - 4. (2) 3s + 4y - 6, 2x - y - 2. (3) VZx + y - 4, V3s - y + 4 = 0. 66. Determine the value of m in y = ma; + 6, so that it shall make an angle of 60° with x - 2y = 3. 67. Find the coordinates of the point through which the three lines y — 4x - 5, y ] — 3s - 4, and y — 2s = 3 pass. Digitized by Google §70] EQUATION OF THE FIRST DEGREE 85 68. Find the value of m so that y — mx + 3 shall pass through the intersection of y — z = 1 and y — 2x = 2. 69. Find the equation of the line perpendicular to 5x -f 8y — 3 and having a y-intercept equal to 6. 60. Find the angle which the line 4a; — y - 8 makes with the line &r - y - 9. 61. Find the equation of the locus of a point whose distance from 3x -f 4y s 5 is one-half its distance from 12s — 5y — 16. 62. Given the two fixed points Pi(-2, 4) and Pj(l, 3). Find the equation of the locus of the variable point P(x, y) which moves so that the area of the triangle PP\P% is always equal to 10. 63. Find the equation of the locus, of a point which moves so that the slope of the line joining it to the point (0, 2) is twice the slope of the line joining it to the point (0, —2). 64. If the equations of the sides of a triangle are x + 2y — 15-0, 2x — y + 5 = 0, and 2x — lly + 15 = 0, find the coordinates of the point of intersection of the bisectors of the interior angles of the triangle. 66. Find the equation of a line passing at a distance y/2 from the origin if the sum of its intercepts is 4. 66. If the three lines A x x + B x y + Ci - 0, A t x + B 2 y + C, + 0, A,x + B*y + C t = 0, meet in a point, show that Ai Bi C x A t B % C 2 = 0. A z Bi C t Digitized by Google CHAPTER V THE CIRCLE AND CERTAIN FORMS OF THE SECOND DEGREE EQUATION 71. Introduction. — The circle affords other examples of the ease and power obtained in analytic geometry by applying algebra to geometry. Since the properties of the circle are well known from plane geometry, atten- tion can be confined to the methods >P(*jt) used in solving the various problems. 72. Equation of circle in terms of center and radius. — A circle is defined + x in plane geometry to be the locus of all points in a plane equidistant from a fixed point in the plane called the center Fio. 68. °f the circle. Let the center of the circle be the fixed point, C(h, k), Fig. 58, and let the constant distance, or radius, be r. Then if P(x, y) is any point on the circle, the distance PC - r. But by [3], PC - V(g - hy + (y - k) \ Then \/(x - h)* '+ (y - k) 2 = r. [25] .\(x-h)* + (y-k) 2 = r 2 . Furthermore, comparison of this equation with [3] shows that every equation of the form of [25] is the equation of a circle. If the center of the circle is the origin, this equation takes the simple form [26] x 2 + y* = r 2 . Digitized by Google §73] THE CIRCLE AND CERTAIN FORMS 87 73. General equation of the circle. — Equation [25] when expanded becomes x 2 + y 2 - 2hx - 2ky + h 2 + k 2 - r 2 - 0. This is in the form [27] x 2 + y* + Dx + Ey + F = 0. \S This is called the general equation of the circle. Conversely f every equation in the form of [27] is the equa- tion of a circle, since after completing the squares in the x and the y-terms, it can be written in the form 2)2 £2 £)2 £J2 x 2 + Dx + — + y 2 + Ey+ _ . _ + _ _ F| or (* + \D) 2 + (y + PO 2 = {WD 2 + E 2 - 4F) 2 . Comparison of this equation with e quation [25], sh ows that h = -iD, k = -\E, and r = WD 2 + E 2 - 4F. Therefore every equation in the form of [27] is the equation of a circle. If D 2 + E 2 — 4F>0, equation [27] represents the equation of a real circle. If D 2 + E 2 - 4F = 0, the radius of the circle equals 0, and the locus becomes a point. Such a circle is called a null or point circle. If D 2 + E 2 — 4jF<0, the radius of the circle is imaginary and the circle is called an imaginary circle. 74. Special form of the general equation of the second degree. — The equation of a circle is a special case of the most general equation of the second degree in two variables Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. In order that this shall be the equation of a circle comparison with [27] shows that B = and A = C, for then this equa- tion becomes Ax 2 + Ay 2 + Dx + Ey+F = f which can be reduced to [27] by dividing by A. The quantity A cannot be zero, since if it were, this equation would become the equation of a straight line. Digitized by Google 88 ANALYTIC GEOMETRY [575 Example. — Find the codrdinates. of the center, and the radius of the circle 6x» + 5y» + 2x - 3y - 4 - 0. Solution. — Dividing by 5. ** + V* + ix - |y - f « 0. Completing squares *■ + 1* + * + »■ - f v + tH - A + tH + f or (x + *)* + (y - A) 1 - tWt- Comparing this equation with [26], shows that the center has the codrdinates (—J, tV) an( ^ that the radius is equal to VSt = t^tV^S- This problem could also be solved by substituting the values D — $ , J0 — -f , and F » -f , in the formulas of Art. 73. EXERCISES Find the codrdinates of the centers and the radii of the following circles: 1. x* + y* - 2x - 4y - 4 - 0. 2. x* + v* + 4s - 6y + 12 - 0. 3. x* + y* + 12* ' + by + 41 - 0. 4. ** + y* - x - 4y + 2 - 0. 5. 3x* + 3y* - 2* - 4y + 1 - 0. 6. 2x* + 2y* + x + 3y - 5 - 0. 7. 2s* + 2y* + 2x + 6y + 5 - 0. 8. a; 1 + y* - 2os - bay + a* « 0. 9. 2x* + 2y* + 12ax + lOay - a* - 0. 10. 9x* + 9y* - 6aa? + 15ay + 6a* - 0. 75. Equation of a circle satisfying three conditions. — Since both equations [26] and [27] involve three arbitrary constants, the circle is determined if enough geometric or algebraic con- ditions are given to determine the three constants uniquely. There are two methods of procedure. One is to compute the constants in [25] geometrically. That is to say, from the given conditions compute the radius of the circle and the codrdinates of its center, then substitute these values in [25]. Another method is to set up three equations involving h, k 9 and r, or three equations involving D, E, and F, and solve these equations simultaneously. This method is generally more satisfactory, and is illustrated for both sets of con- stants in the following example. Digitized by Google 575] THE CIRCLE AND CERTAIN FORMS 89 Example 1. — Find the equation of a circle passing through the points (3, 5), (4, 4), (1, 1). First method, geometrical. — Find the equations of the perpendicular bisectors of two of the sides of the triangle (3, 5), (4, 4), (1, 1). Solve these equations simultaneously. This gives the coordinates of the center of the circle. Next find the distance from the center of the circle to any one of the three vertices. This gives the radius of the circle. Substituting the values of h, k and r thus found in [26] gives the desired equation. It is obvious that this method is long and hence the actual computa- tion is not given. A shorter method is the following. Second method, algebraic. — Make the coordinates of each of the three points satisfy the equation (x — h) % + (y — *)* ■ **• Tk* 8 8* ve8 (3 _ h) i + ( 5 _ *)» - r«, * (4 - *)* + (4 - *)« - r«, (1 - A)« + (1 - *)* - r*. Simplifying each of these equations gives h* + &* - 6fc - 10* - r« + 34 - 0, h* + h* - 8A - 8fc - r* + 82 - 0, ' A* + *« - 2h - 2* - r* + 2-0. Solve these equations by subtracting the second from the first and the third from the second. Then solving the two equations thus obtained gives h — 2, k = 3, r — y/b. Hence the equation of the required circle is (x - 2)* + (y - 3)* - 5 Simplifying, this becomes x* + y* - 4s - 6y + 8 - 0. Third method. — Make the coordinates of the three points satisfy the equation x % + y* + Dx + Ey + F « 0. This gives 9 + 25 + 3D + SE + F - 0, 16 + 16 + 4Z> + 4E + F - 0, Solving these equations simultaneously gives D - -4, JE - -6, F - 8. Hence the equation of the circle is x s + y»- 4x - 6y + 8 - 0. Example 2. — Find the equation of a circle which passes through the points ( — 1, 7) and (7, 1) and is tangent to the line x + y — 10 - 0. Digitized by Google 90 ANALYTIC GEOMETRY [§75 This problem illustrates how a combination of both algebraic and geometric methods may sometimes be useful. Solution. — Use the equation (x — A) 1 + (y — k)* — r f , and make the circle go through the points ( — 1, 7) and (7, 1). This gives the two equations (-1 - A)« + (7 - k)* » r«, (1) and (7 - A)« + (1 - ib)« =r«. (2) Since the line x + y — 10 = is tangent to the circle, the distance from the point (A, k) to the line x + y — 10 « equals r, hence by [23] Y A + k - 10 \ I v^foS (Wk X w tr o \5 (3) like ±V2 Simplifying and combining terms in (1) and (2) gives A* + ib* + 2A - 14& + 50 « r*, (4) A* + fc* - 14A - 2fc + 60 - r*. (5) Subtracting (5) from (4) and divid- ing both sides of the resulting equa- tion by 4, 4A - Zk - 0. (6) Substituting the value of r from equation (3) in (4) and simplifying, A* - 2hk + k* + 24A - 8k - 0. (7) Substituting A = £ k from equation (6) in equation (7) and simplifying, k* + 160fc = 0. Hence k = or h = -160. Computing the value of A from equation (6), gives A = or A — — 120. Computing the value of r from equation (3), gives r = 25\/2 orr = 145\/2. Substituting the values of A, &, and r in the general equation of the circle gives the two solutions x* + y* — 60, Fig. 59. and (x + 120)* + (y + 160)* - 42,050. EXERCISES Find the equations of the circles through the following points: 1. (5, 5), (1, 3), (2, 6). 2. (3, -2), (-1, -4), (2, 3. (0,3), (-4,3), (-3,4). 4. (1,6), (4,5), (-3, -2). 5. (-1,4), (-4, -5), (3,2). 6. (2,8), (-1, -1), (-2,6). 5). 7. (2, 4), (2, -2), (3, 3). 8. (4,4), (5,3), (-3,3). 9. (3, 7), (1, 1), (5, 3). 10. (-1,1), (1,5), (-5,3). Digitized by Google §75] THE CIRCLE AND CERTAIN FORMS 91 Find the equations of the circles fulfilling the following conditions:' 11. Passing through the origin, radius 5, and ordinate of center —3. 12. Passing through the origin, radius 13, and abscissa of center 12. 13. Center at origin and tangent to line x + 2y — 10. 14. Center at point (1, 2) and passing through the point (3, — 1). 16. Center at ( — 1, 3) and tangent to line Zx + y — 10 — 0. 16. Center on x-axis and passing through the points (3, 3) and (5, —1). 17. Radius 5 and passing through the points (5, 6) and (2, 7). 18. Radius 5 and tangent to the line 4x + Zy «- 16 = at the point (1, 4). 19. Having the line joining (—3, 2) and (5, 6) as diameter. 20. Passing through the point (1, 1) and having the same center as x s + y % + 4s - 6y - 0. 21. Intercept on x-axis equals 3, and passing through the points (-1,2) and (2, 3). 22. Tangent to x-axis, radius 4, and abscissa of center 3. 23. Tangent to y-axis, radius 2, and ordinate of center 4. 24. Center on the line x — y + 2 = 0, and passing through the points (3, 7) and (1, 1). 26. Center on the line 2x — y — 3 = 0, tangent to both axes, and in the first quadrant. 26. Center on the line 2x — y — 3 = 0, tangent to both axes, and in the fourth quadrant. 27. Center on the line 3x — y + 8 = 0, tangent to both axes, and in the. second quadrant. 28. Radius 3, tangent to both axes, and in the second quadrant. 29. Tangent to the line 3a; + y -f- 2 « at the point (-1, 1) and passing through the point (3, 5) 30. Intercept on the y-axis 4, and tangent to the line x + 2y + 1 — at the point (—3, 1). 31. Tangent to both axes, in the second quadrant, and also tangent to the line 3x — 4y + 30 — 0. (Two solutions.) 32. Tangent to both axes, in the first quadrant, and also tangent to the line 3x - 4y + 30 = 0. 33. Tangent to both axes and passing through the point (8, 1). (Two solutions.) 34. Find the equation of the diameter with slope 2 of the circle x* - 4* + y* + 6y - 3 = 0. 36. The point ( — 1, 2) bisects a chord of the circle x* + y* - 10. Find the equation and length of the chord. 36. A chord of the circle x* + y* + 2x + 4y - 15 = is bisected by the point (—2, 1). Find the equation and length of the chord. Digitized by Google 92 ANALYTIC GEOMETRY [§76 37. Find the equation of the circle inscribed in the triangle whose sides are the lines 6a; + 7y « 86, — 7x + 6y « 86, and 2x — 9y — 86. 88. Find the equation of the circle inscribed in the triangle whose sides are the lines Zx 4- 4y » 18, —4a; + 3y » 26, and y + 4 — 0. 89. Find the equation of the circle circumscribing the triangle whose sides are the lines 7a; + 9y - 66, 3a; + V — 26, and x + 2y « 16. 40. Prove analytically that an angle inscribed in a semicircle is a right angle. 41. Prove analytically that a line from the center of a circle bisecting a chord is perpendicular to it. Suggestion. — Let the ends of the chord be (r, 0) and (6, c). 42. Prove analytically that the length of a perpendicular from any point on the circumference of a circle to a diameter, is a mean propor- tional between the segments into which it divides the diameter. 48. Prove that the length of the tangent from the point (xi, y\) to the circle x* + y* + Dx + Ey + F - is x x * + y x * + Dxi + Ey x + F - 0. 76. Systems of circles. — If fi(x, y) = and f 2 (x, y) = are the equations of any two circles, then by article 67 /i(s> y) + kf 2 (x, y) = is the equation of a curve through all the points of intersection of fi(x, y) = and f 2 (x, y) = 0. Furthermore in this case the curve will always be a circle or a straight line. To prove that this is so, let/i(», y) = stand for the equa- tion Aix 2 + Aiy 2 + DiX + Exy + Ft = 0, and let /,(*, y) = stand for A& 2 + A%y s + D2X + E 2 y + F % = 0. Then f 1 (x f y) + kf i (x f y) =0 becomes A\x* + A x y % + D x x + Eiy + Ft + k(A&* + A t y* + D2X + E iy + F 2 )=0. Collecting like powers of x and y, this equation becomes (Xi + kA 2 )x* + (At + kA 2 )y* + (D x + kD t )x + (ft + *ft)y + ft + Aft = 0. Since the coefficient of x 2 equals the coefficient of y 2 and the coefficient of xy equals 0, this is the equation of a circle. The Digitized by Google 576] THE CIRCLE AND CERTAIN FORMS 93 exception occurs when Ai + kA* ■» 0, in which case, this equation is of the first degree and therefore is the equation of a straight line. Fio. 61. Example 1. — Find the equation of a circle through the point (1, 2) and the points of intersection of the circles 2s* + 2y* — 3s — 4y — 1 - 6 and 3s* + 3y* - 8s - y - 4 - 0. Solution. — The equation of any circle through the points of intersection of these two circles is 2s* + 2y* - 3s - 4y - 1 + fc(3s* + 3y* - Sx - y - 4) - 0. Since the point (1, 2) is on this circle its coordinates must satisfy the equation of the circle, therefore 2+8-3-8-1+ &(3 + 12 -8-2- 4) =0. Solving for k, gives k « 2. Therefore the required equation is * 2s* + 2y* - Zx - 4y - 1 + 2(3s* + 3y* - Sx - y - 4) - 0, or Sx* + Sy* - 19s - 6y - 9 - 0. Example 2. — Find the equation of the common chord of the circles, 2s* + 2y* - 6s - 4y + 1 - and s* + y* - 2s - y + 3 - 0. Solution. — The equation of any circle through the points of inter- section of these two circles is 2s* + 2y* - 6s - 4y + 1 + fc(s* + y* - 2s - y + 3) - 0. In order that this equation shall be the equation of a straight line, it is necessary that the coefficient of s* shall vanish, hence 2 + h = 0. This gives A; — —2 Digitized by Google 94 ANALYTIC GEOMETRY [§77 Making this substitution the equation becomes 2s* + 2y* - 6s - 4y + 1 - 2(s* + y* - 2s - y + 3) - 0, or 2s + 2y + 5 - 0. This is the equation of their common chord. If the two circles intersect in real points, the straight line thus obtained is their common chord, since it passes through their two points of intersection. If the two circles do not intersect visually, they are still said to intersect algebraically, their points of intersection being imaginary, and the line /i (z, y) + kfz (x, y) passes through their imaginary points of intersection. The straight line which passes through the real or imaginary points of intersection of two circles is called their radical axis. EXERCISES Find the equation of the common chord or the radical axis of the circles in exercises 1-6. l.x , + l/ , -3x+ y-6=0, 4. 2s* + 2y* - 3s - Zy + 5 - 0, s* + y* - 5s - Zy + 4 = 0. 3s* + Zy 1 - 2x - Zy + 4 - 0. 2. s* + y* - 6s - Sy + 3 = 0, 6. 4s* + 4y* - x + y - 6 - 0, x s + y* + 4s + 2y - 7 - 0. 3s* + 3y* - 2x - 3y + 4 - 0. 3. x* + y* - 3s - 4y + 2 - 0, 6. 3s* + 3y* - 2s - Zy + 6 - 0, s* + y* - 2s - 2y + 6 - 0. 2s* + 2y* + s + y - 2 = 0. 7. Find the equation of the circle through the point (1,1) and through the points of intersection of the circles s* + y* - 2s - Zy + 4 - 0, s* + y* - 4s - by + 6 - 0. 8. Find the equation of the circle through the point (3, 4) and through the points of intersection of the circles s* + y* - 7s - Zy + 10 - 0, s* + y* - 8s + 2y - 6=0. 9. Prove that the common chords of the following circles, taken two at a time, meet in a point: s* + y* - 4s - Zy + 6 = 0, s* + y* - 2s + by - 2 - 0, s* + y* + s + 2y - 4 - 0. 77. Locus problems involving circles. — Although the ele- ments dealt with in plane geometry are the point, straight Digitized by Google §77] THE CIRCLE AND CERTAIN FORMS 95 Y \ J>f f * ,lf) ^)J 7a4 /r(24) I \ ' ° — i 1 i — 4**x line and circle, nevertheless the locus problems that can readily be handled by plane geometry are only of the simplest kind. On the other hand analytic geometry lends itself easily to the solution of locus problems as is illustrated by the following example. Example. — Find the locus of the point, which moves so that the sum of the squares of its distances from the points (0, 1) and (2, 1) is constant and equal to 20. Solution. — Let P(x, y) be any point ' on the locus, then FS* + FT* = 20, (1) FS* - x* + (y - l)*, FT 1 - (x - 2* + (y - 1)*. Substituting these values in equa- tion (1) x* + (y - 1)» + (x - 2)» + (y - 1)* Simplifying, a; 2 + y* — 2x — 2y = 7. Completing the squares in the x and y-terms, (x - 1)» + (y - 1)« - 3*. Hence the required locus is a circle whose center is the point (1, 1) and whose radius is 3. EXERCISES 1. Find the locus of a point which moves so that the sum of the squares of its distances from (—2, 0) and (2, 0) is constant and equal to -26. 2. Find the locus of a point which moves so that the sum of the squares of its distances from ( — 1, 2) and (2, 1) is constant and equal to 10. 3. Find the locus of a point such that its distance from the point (—2, 0) shall always be twice its distance from the point (2, 0). 4. Find the locus of a point moving so that its distance from the line Zx + 4y — 5 =0 shall equal the square of its distance from the point (1, 0). 6. Find the locus of a point such that its distance from the y-axis shall equal the square of its distance from the point (0, 2). (Two solutions.) 6. In an isosceles triangle of base 6 and equal sides of length 5, a point moves so that the product of its distances from the equal sides equals the square of its distance from the base. Prove one of the loci to be a circle and find its radius. Fig. 62. 20. Digitized by Google 96 ANALYTIC GEOMETRY [§78 7. Find the locus of the vertex of a right angle if its two sides always pass through the points (—2, —4) and (2, 6). 8. Find the locus of the vertex of an angle of 30°, whose sides pass through the points (—2, 0) and (2, 0). (Two solutions.) 9. Find the locus of the vertex of a triangle, if the remaining two vertices are at the points (—3, 0) and (3, 0) and the length of the median from the vertex ( — 3, 0) is constant and equal to 5. 10. The ends of a straight line of length 6 rest on the axes, find the locus of its middle point. 78. Equation of a circle in polar coordinates. — Let the radius of the circle be r, and let C(pi, 0i) be the coordinates of its center, Fig. 63. Then if P(p, 0) is any point on the Jr^^l^ circle, by trigonometry, r 2 = p 2 + Pl 2 - 2p Pl cos COP. Replacing angle COP by its value (0 - 0i), r 2 = p 2 + Pl 2 - 2 PPl cos (0 - 0i). "*" The forms of this equation which Fiq. 63. occur most frequently are those where the center is the pole or where the circle passes through the pole and the center of the circle is either on the initial line or on the line = 90°. If the center is the pole, pi = 0, and the equation becomes p = r. If the circle passes through the pole and has its center on the initial line, 0i = and p\ = ±r. The equation of the circle then becomes p = 2r cos 0, or p = — 2r cos 0, according as the center is on the initial line or the initial line produced through the pole. If the circle passes through the pole and its center is on the line = 90°, B\ = 90° and p = ±r, and the equation becomes p = 2r sin 0, or p = — 2r sin 0, according as the circle lies above or below the polar axis. Digitized i by Google §78] THE CIRCLE AND CERTAIN FORMS 97 EXERCISES Find the equations of the following circles in polar coordinates: 1. The center is at the pole and the radius equals 2. 2. The center is at the point (5, 0) and the radius equals 5. 3. The center is at the point (—4, 0) and the radius equals 4. 4. The center is at the point (3, far) and the radius equals 3. 6. The center is at the point (—2, far) and the radius equals 2. 6. The circle is tangent to the initial line at the pole and the radius equals 6. 7. The circle is tangent to the line $ » 90° at the pole and the radius equals 6. 8. The center is at the point (3, 1*-) and the radius equals 3. Change from rectangular to polar coordinates. 9. x* + #* - 6. 11. 2x* + 2y* + 5x - 0. 10. x* +-y* - 3y - 0. 12. x* + y* - Ox - 8y - 0. Change from polar to rectangular coordinates and find the center and radius of each of the following circles. 13. p + 6 sin $ = 0. 17. p + 2 cos $ + 3 sin $ - 0. 14. p - 4 cos $ = 0. 18. p* + 3p cos $ + 4p sin $ - 6 - 0. 16. p = cos $ + sin $. 19. p* = 9 sec*0 — p f tan*0. 16. p = 5. 20. p f - 4 cscf - p* cot»0. Digitized by Google CHAPTER VI THE PARABOLA AND CERTAIN FORMS OF THE SECOND DEGREE EQUATION 79. General statement. — It is an interesting and useful fact that an equation of the second degree in two variables, if plotted with reference to rectangular axes, gives a conic 8ection y or simply a conic. That is, the graph is some plane section of a right circular cone. 80. Conic sections. — When a plane intersects a circular cone there may be formed a circle, a parabola, an ellipse, an hyperbola, or, for cer- tain positions of the plane, a point, two in- tersecting straight lines, or two coincident lines. In Fig. 64, plane C is perpendicular to the axis of the cone and forms a circle; plane disinclined to the axis but intersects only one nappe of the cone and forms an ellipse; plane P is par- allel to an element of plane H intersects both The intersec- Fio. 64. the cone and forms a parabola; nappes of the cone and forms an hyperbola. tion is a point when a plane passes through the point V only; two intersecting straight lines are formed when the plane passes through V and intersects the nappes; and two Digitized by Google §81] THE PARABOLA AND CERTAIN FORMS 99 coincident lines are formed when the plane passes through V and is tangent to the cone. The conic sections were first studied by the Greeks, who discovered and discussed their properties by methods of geometry. The modern method of studying these figures is by the help of algebra, which makes the treatment much simpler. For the purposes of this method of treatment, other definitions of the conic sections are given; but it can be readily shown that these definitions agree with the definitions mentioned above. EXERCISES 1. Explain how a conic section could be two lines inclined to each other at an angle of 45°. Could the two straight lines formed on the same cone form different angles with each other? 2. If the vertex angle of a cone is 30°, what would be the angle between the intersecting lines formed by the plane intersecting the cone? 3. In forming an hyperbola, does the plane have to be parallel to the axis of the cone? Could hyperbolas of different shapes be formed on the same cone? 4. Explain how a parabola of different widths could be formed on the same cone. 5. Explain how ellipses of different widths could be formed on the same cone. Explain the change in the shape of the ellipse formed by a plane that revolved into a position parallel to an element of the cone. 81. Conies. — A definition of a conic section, and one that can readily be translated into algebraic language, is the following: A conic is the locus of a point that moves in the plane of a fixed straight line and a fixed point not on the line, in such a manner that its distance from the fixed point is in a constant ratio to its distance from the fixed line. The fixed point is called the focus of the conic, and the fixed line is called the directrix. The constant ratio is called the eccentricity and is usually represented by e. . < The constant e is positive, and may be equal to 1, less than 1, or greater than 1. Digitized by Google N <-H p.v) N' V 2 pO 100 ANALYTIC GEOMETRY [§82 If e = 1, the conic is a parabola. If e < 1, the conic is an ellipse. If e > 1, the conic is an hyperbola. 82. The equation of the parabola. — By the definition of the preceding article, the parabola is the locus of a point equidistant from the focus and the directrix. In Fig. 65, let F be the focus and D'D the directrix. Choose as x-axis the line X'X through F and perpendicular to D'D p(*.v) at R. The point on X'X midway between R and F is a point on the locus. Choose this point as origin. Then —*x Y'Y parallel to D'D is the y-axis. Let p represent the length and direction of RF. Then the coordinates of F are ($ p, 0), and the equation of D'D is x = -£p.. To derive the equation of the parabola, let P(x, y) be any point on the locus, and draw FP, and NP perpendicular to D'D. By definition FP = NP. But FP = V(x - £p) 2 + y\ and NP = x + \p. Then V(* - \pY + y 2 - * + ip. Squaring and simplifying, this becomes [28] iyj = 2px. N > The simple form of this^ equation is due to the choice of the coordinate axes. If they had been chosen differently, the equation would be more complicated; but the locus itself would be unaltered. Equation [28] is the required equation. For it has been proved true for every point on the parabola; and it is not true ViV FiViP.Q) Y c Fio. 65. Digitized by Google 583] THE PARABOLA AND CERTAIN FORMS 101 for any point that is not on the parabola, for then FP is not equal to NP, and therefore y 2 is not equal to 2px. It should be remembered, that in the equation y 1 = 2px, p represents the length and direction of RF. Therefore, when the focus lies to the right of the directrix, p is positive; but, when the focus lies to the left of the directrix, p is negative. 83. Shape of the parabola. — The shape of the parabola and its position relative to the coordinate axes can be readily determined from the equation y 2 = 2px. Solving for y gives y = ±\/2px. For any positive value of p we have: (1) When x = 0, y = 0. Hence the curve passes through the origin. (2) For all positive values of x, y has two numerically equal values but opposite in sign. Hence' the curve is symmetrical with respect to the x-axis. (3) For any negative value of x, y is imaginary. Hence nd part of the curve is at the left of the y-axis. As x in- creases from 0, the positive value of y increases and the negative value decreases. The curve can be located more precisely by the following points: Fig. 66. X \v P 2p 4p Sp 50p y ±p ±py/2 ±2p ±2pV2 ±4p ±10p The parabola has the shape shown in Fig. 66. It is evident that all parabolas have the same shape, the appearance digitized *by Google 102 ANALYTIC GEOMETRY 1184 depending only upon the sue of the unit chosen. For a negative value of p, the parabola will be exactly the same shape but opening toward the left. 84. Definitions, — The point of the parabola midway between the focus and the directrix is called the vertex of the parabola. The line through the focus and perpendicular to the directrix is called the axis of the parabola. As has been proved in the preceding article, the axis bisects all the chords of the parabola which are parallel to the directrix, once the axis of the para- bola lies on the £-axis. The chord of the parabola through the focus and perpen- dicular to the axis is called the latus rectum. The length of FiO. 67. »*• 2 PV , V positive x*m 2py . v negative Fio. 68. the latus rectum is the absolute value of 2p. For the ab- scissa of the focus is Jp, and, when « » |p, y =r ±p m In Fig. 67, V is the vertex of the parabola, VX is the axis, and P'P is the latus reetum. A parabola can be readily sketched if the position of the vertex V and focus F and the length of the latus rectum, P'P, are known. 85. Parabola with axis on the y-axis. — The equation of a parabola whose axis is on the y-axis and whose vertex is at the origin is obviously obtained by interchanging x and y in the work of article 82. The equation is 120] 2py. Digitized by Google §861 THE PARABOLA AND CERTAIN FORMS 103 The focus is at the point (0, Jp), and is on the positive or the negative half of the y-axis according as p is positive or negative. If p is positive, the parabola, Fig. 68, is above the x-axis; and, if negative, it is below the x-axis. It is to be remembered that the origin is at the vertex of every parabola whose equation is of the form [28] or [29]. These forms are called the standard forms of the equation of the parabola. EXERCISES 1. Plot the following parabolas: y* = 2x t y* = — 2x t x* = 2y t and 2. Give the coordinates of the foci of the parabolas in exercise 1. Give the equations of their directrices. What are their latera recta? 3. Plot y* = 4x, using successively A hi., i in., J in., i in., 1 in., and 2 in. as a unit. 4. Plot y* — ix using 4 in. as a unit. Plot y* « \x using 1 in. as a unit. Plot y* — x using i in. as a unit. Plot y* « 4x using J in. as a unit. Are all parabolas of the same shape? 5. Write the equation of a parabola whose vertex is at the origin and focus at (1) (3, 0), (2) (0, 6), (3) (-4, 0), (4) (0, -2). 6. Find the equations of the following parabolas, and give the latus rectum of each: (1) Vertex at origin, axis on x-axis, and passing through the point (2, 4). (2) Vertex at origin, axis on y-axis, and passing through the point (2, 4). 7. The cables of a suspension bridge hang in the form of a parabola. Find the equation for such a cable in a bridge 1000 ft. between supports if the distance from the lowest point of the cable to the level of the top of the piers is 50 ft. Suggestion. — Take the origin at the lowest point of the cable. Then the point (500, 50) is on the parabola. Substitute these values in [29] and solve for p. 8. Derive equation [29] from [28] by revolving the coordinate axes through an angle <p = —90°. 86. Equation of parabola when axes are translated. — Transform the equation y 2 = 2px by translating the axes to anew origin at the point 0'(— h, —A), Fig. 69. Digitized by Google 104 ANALYTIC GEOMETRY [§86 Y' ->X By [12], x = x' — h and y = y' — fc. Substituting these values in y 2 = 2px gives (y / - fc ) 2 = 2p(x' - A). This is the equation of a parabola having its vertex at the point (h, k) when referred to the new coordinate axes, that is, the x*, y'-axes. If the primes are dropped, this becomes [30] (y - k)* = 2p(x - h), which is a convenient form for writing the equation of a para- bola with vertex at point (h, k) and axis parallel to the 3-axis. If p is positive, the parabola opens toward the right; and if negative, it opens toward the left. Similarly, when the axis of the parabola is parallel to th£ y-axis, the equation is [30x] (x - h) 2 = 2p(y - k). (-A.-*> -+X' Fio. 69. *»X (1) \^ (2) p positive p negative (3) / (1) p pbsitive p negative Fio. 70. The position of these parabolas with reference to the coordinate axes is shown in Fig. 70. Example 1. — Find the equation of a parabola with vertex at the point (2, —3), axis parallel to the a>axis, and p = 2. Plot. Substituting in [30], (y + 3) 2 = 2 X 2(x - 2). Simplifying, y* + 6y - 4s + 17 = 0. The curve is plotted in Fig. 71. Digitized by Google §861 THE PARABOLA AND CERTAIN FORMS 105 Example 2. — Fine} the equation of the parabola whose vertex is at the point (3, —6), axis parallel to the y-axis, and which passes through the point (-3, -10). Solution. — The equation is of the form Y [30i], in which h — 3, A; — —6, and p is to be found. x Substituting in [30i], (_3 - 3)*«2p(-10, +6). Solving for p, p — —4$. Substituting values of h, k, and p in [80i] gives (*-3)*«2(-4i)(y + 6). Simplifying, x* - 6x + 9y + 63 = 0, r' the required equation. Fia. 71. EXERCISES 1. Write the equations of the following parabolas: (1) Vertex at (3, 4), p =4, and axis parallel to the re-axis. (2) Vertex at (2, 3), p = —4, and axis parallel to the x-axis. (3) Vertex at (— 6, 2), p «■ 6, and axis parallel to the s/-axis. (4) Vertex at (2, — 3), p = —3, and axis parallel to the y-axis. 2. In each part of exercise 1, give the codrdinates of the focus, equa- tion of the directrix, and plot the parabola. 3. Write the equations of the parabolas with vertex of each at (—4, —2), latus rectum of each equal to 10, and axes parallel to x-axis. 4. Write the equation of the parabola with vertex at (3, —2), origin on the directrix, and axis parallel to y-axis. 5. Transform x* + By = 12 to new axes parallel t* the old, with the new origin at the point (2, 5). 6. Find the equations of the following parabolas, and sketch each curve: (1) Vertex at (4, 5) and the focus at (6, 5). (2) Vertex at (-4, 2) and the focus at (-4, 4). (3) Vertex at (-4, 2) and the focus at (-6, 2). (4) Vertex at (3, -4) and the focus at (3, -6). 7. Find the equations of each of the following parabolas, and sketch each curve: (1) Vertex at (2, 3), axis parallel to x-axis, and passing through the point (5/6). (2) Vertex at (3, —2), axis parallel to x-axis, and passing through the point (-1, 3). Digitized by Google 106 ANALYTIC GEOMETRY [§87 I (3) Vertex at (2, 3), axis parallel to y-axis, and passing through the point ( — 1, 1). (4) Vertex at (3, —2), axis parallel to y-axis, and passing through the point (-1, 3). 87. Equations of # forms y 2 + Dx + Ey + F = <ah!i x 2 + Dx + Ey + F = *0. — (1) Every equation of the form y 2 + Dx + Ey + F = 0, where D j* 0, represents a parabola whose axis is parallel to the x-axis. (2) Every equation of the form x 2 + Dx + Ey + F = 0, where E j* 0, represents a parabola whose axis is parallel to the y-axis. Proof of (1).— Given y 2 + Dx + Ey + F = 0, where Z)>* 0. E 2 E 2 -rH Completing square in y, y 2 + Ey + -r- = — Dx + -j- — F. Or / , E\ 2 n/ E 2 -4F\ •'••'♦ (y + 2) - - D (* --&-)• E 2 — 4F E This is in the form of [30], where h = — j~ — , k = — ^ A D U and p « -^ Therefore the equation j/ 2 + Z)s + Ey + F = where D j* represents a parabola whose axis is parallel to the x-axis. The proof of (2) is similar to that of (1). Example 1. — Transform the equation y* + 4y — 4s + 8 =0 into the form of [30], give the coordinates of the vertex x and focus, write the equations of the axis and directrix, and sketch the parabola. Solution.-— Completing the square in y, y* + 4y + 4 = 4x - 8 + 4. Or (y + 2)» - 4(s - 1). Hence the vertex is at the point V (1 - 2). Since 2p = 4, p = 2, and the focus is Fio. 72. one unit to the right of the vertex, or at the point (2, -2). The axis is parallel to the x-axis and two units below. Hence its equation is y = — 2* Digitized by Google §88] THE PARABOLA AND CERTAIN FORMS 107 The directrix is perpendicular to the x-axis and one unit to the left of the vertex. Hence its equation is x - 0. The parabola is shown in Fig. 72. Example 2. — Find the equation of the parabola with its axis parallel to the x-axis, which passes through the points (0, 1), (2, 3), (5, 2). Solution. — The equation is of the form y* + Dx + Ey + F - 0. Since the parabola passes through the point (0, 1), these coordinates satisfy the equation. Substituting these coordinates gives 1 + E + F - 0. Likewise (2, 3) give 9 + 2D + SE + F - 0. And (5, 2) give 4+ 52> + 2# + F « 0. Solving these equations for 2>, E, and F, D - J, E - — ¥, and F - *f. Substituting these values in y» + Dx + Ey + F - 0, gives y 1 + \x - V* + ¥ - 0. Or 4y* + x — 17y + 13 — 0, the required equation. , 88. The quadratic function ax 2 + bx + c. — The locus of the equation y = ax 2 + bx + c, where a, 6, and c are real numbers and a 5*0, is a parabola with axis parallel to the y-axis. To see this, reduce the equation to the standard form [30J. b 2 — 4oc / b\ 2 Completing the square in x> y = o ( x + =- ) 4 -_4oc\ 4a / / « & \ 2 1 / , &* - 4ac \ This is in the form (x — h) 2 = 2p(y — fc), where h - — =- and fc = j > and is a parabola with vertex at the point (— «- * Z ) anc * ax ^ on ***© ** ne x "t" 9~ "" ®' Evidently the parabola opens upward if a > and downward ifa<0. 89. Equation simplified by translation of coordinate axes. — It is evident that y 2 + Dx + Ey + F = and x 2 + Dx + Ey + F = can be transformed to the forms of Digitized by Google 108 ANALYTIC GEOMETRY I§89 [28] and [29], respectively, by a suitable translation of the coordinate axes. . In the first equation, the term in y and the constant term can be made to vanish; and, in the second, the term in x and the constant can be made to vanish. Example. — Translate the coordinate axes so as to transform the equation y* + 6s — 4y + 10 = to the form of y* = 2px. Solution. — Using [12], put x — z? + h and y — y' + k, then (y' + *)» + 6(x' + h) - My* + *) + 10 = 0. Or y 7 * + 6V + (2fc - 4)y> + (A; 1 - 4* + 6h + 10) - 0. In order that the y' term and the con- stant term shall vanish 2fc - 4 = and h* - 4fc + 6h + 10 - 0. Solving these equations, h = — 1 and fc = 2. Therefore the transformed equation is y** - -6*'. The transformation can also be made by completing the square in y, whence y* - 4y + 4 = -6s — 6, or (y - 2)» = -6(s + 1). Put y — 2 = y' and 3 + 1 - s', and obtain y' 1 = — 6s', as before. The curve is plotted in Fig. 73. *~x +-X Fio. 73. EXERCISES 1. Transform the equations of the following parabolas to the form of [30] or [30 J; and in each case give the coordinates of the vertex and the focus, write the equations of the axis and directrix, and plot. (1) y* - 4x - 4y + 16 ~ 0. (4) 2x* - 24x + 3y + 78 = 0. (2) y 1 + 2x + 8y + 6 = 0. (5) 3y» + lfo' - 12y + 20 = 0. (3) 4x* + \2x - 20y + 49 = 0.- (6) 2x* - 18s + 15y - 21 - 0. 2. Find tfce equation of the parabola with axis parallel to the y-axis, which passes through the points (2, 3), (1, 0), and (0, 2). Find the coordinates of the focus and vertex of this parabola, and its latus rectum. 3. Find the equation of the parabola which has the line y = 4 as axis, the line x = — 2 as directrix, and p = 6. 4. Find the equation of the parabola which has its vertex at (2,-3), its axis parallel to the z-axis, and which passes through the point (5, 2). Digitized by Google §90] THE PARABOLA AND CERTAIN FORMS 109 5. Translate the coordinate axes so as to transform the following parabolas to the form of [28] or [29]. In each case plot showing both sets of axes. (1) y* - 4x - 6y + 8 - 0. (3) y* + 8x - 4y - 4 - 0. (2) x* - 8x + 16y - 0. (4) 3x» + 6x - 7y + 8 - 0. 6. For each of the parabolas of exercise 5, find the equation of the directrix with reference to both sets of axes. Give the coordinates of the focus for both sets of axes, and the value of the latus rectum. 7. Plot the equation y — ax 1 + bx + c discussed in article 88 for (1) &» - 4oc > 0, (2) 6* - 4ac « 0, (3) 6* - 4ac < 0, both when a > and when a < 0. 90. Equation of a parabola when the coordinate axes are rotated. — Transform the equation y 2 + Dx + Ey + F = by rotating the coordinate axes through an angle <p, using the formulas [13]. Putting x = x' cos <p — y' sin <p, and y = x f sin <p + y 9 cos <p, in y 2 + Dx + Ey +F = 0, gives (a/ sin ¥> + !/' cos *>) 2 + Z)(a/ cos <p — j/' sin ^) + E(x' sin ¥> + y' cos <p) + F = 0. Collecting terms, x' 2 sin 2 ip + 2 sin ¥> cos <p x'y' + y' 2 cos 2 <p + (D coa <p + E sin *>)z' + (E cos <p-D sin *>)y'+F = 0. (I) A similar form is obtained from x 2 + Dx + Ey + F = 0. If the angle of rotation is some multiple of 90°, then 2 sin <p cos <p = 0, and the coefficient of x'y' is 0. Hence, in this case, the x'y'-tevm vanishes. If the coordinate axes are rotated through an angle <p f such that the axis of a parabola is not parallel to either coordinate axis, the equation of a parabola is of the form Ax 2 + Bxy + Cy 2 + Dx + Ey+F = 0, (II) the most general form of an equation of the second degree in x and y. » It is readily seen that in equation (I), B 2 — 4AC = 0. It will be shown later, Art. 121, that the necessary and sufficient condition that any equation of the form of (II) represents a parabola is that B 2 — 4AC = 0. Digitized by Google 110 ANALYTIC GEOMETRY [§90 Example 1. — Transform the equation x % + 2x — 3y + 4 — by rotating the coordinate axes through an angle of 45?. Plot. Solution. — Substituting x — x' cos 45° — y' sin 45° and y — x 9 sin 45° + j/ 7 cos 45°, (a/ cos 45° - y' sin 45°)* + 2(x / cos 45° - tf sin 45°) - 3(0/ sin 45° + y> cos 45°) +4=0. Simplifying, *>* - 2*V + y 71 - V2*' - 5 W +8 = 0. Example 2. — By rotating the coordinate axes transform the equation 9x* -^i^+ 16V — 116x - 162y + 221 = 0, to a form which contains no term m xy. Solution. — Putting x = a/ cos *> — y* sin ?, and y = a/ sin <p + y* cos <p, 9(x / cos *> — y' sin ^) s — 24(s' cos tp — y' sin ^Xs' sin ? + y' cos *>) + 16(3' sin *> + y 7 cos *>)* — 116(:r/ cos tp — y' sin *>) — 162(x / sin ? + y> cos <p) + 221 = 0. Collecting terms, (9 cos 1 *>— 24 sin ? cos *> + 16 sin 1 <p)x* % + • (14 sin ^> cos ^> + 24sin 1 *> — 24cos , *>)x / y / + (9sin , ^ + 24sinv>coS^ + 16 cos 1 *>) y' 1 — (162 sin *> + 116 cos <p)x' + * (116 sin *>- 162 cos ^Jy 7 + 221 = 0. Now, in order that the x*\f term shall vanish, its coefficient must be 0. Hence ^24 sin 1 y — 24 cos 1 <p + 14 sin ^ cos y = fl. Or -24 cos 2tp + 7 sin 2? - 0. Dividing by cos 2?, 7 tan 2*> = 24, or tan 2<p - *#>. From this by trigonometry, cos 2? = /j. Then And — cos 2 COS <p + cos 2^ "2~ *' + A f Substituting these values for sin <p and cos *> in the above equation and simplifying, 25^ 1 _ = ^190aj / - 60y' + 221 - 0. EXERCISES 1. Transform the equations y 1 = 2px and x 1 = 2py by rotating the coordinate axes through an angle of 90°. 2. Transform the following equations by rotating the coordinate axes through the angle given in each case: (1) y 1 - Ax. * = 46°. (2) a 1 + 3s - 2y + 6 - 0. = 30°. Digitized by Google §91] THE PARABOLA AND CERTAIN FORMS 111 (3) 4x* - 4xy + y* + 2x - 6y - 10 - 0. ? - sin" 1 iVS. (4) 9s* + 12xy + 4y« + lOx - 64y - 68 - 0. 2* - tan" 1 -^. 3. Derive the equation of the parabola whose directrix is the line 4x +3y + 2 — 0, and whose focus is at the point (2, 3). 4. Simplify the following equations, and plot. First rotate the coordi- nate axes to free of xy-term, then translate to change to the standard form. (1) s 1 - 2xy +ty -6x-6y+9»0. (2) 2x» + Sxy + 8y* + x + y + 3 = 0. (3) x* + 2xy + y* - 12x + 2y - 3 - 0. 91. Equation of parabola in polar coordinates. — Starting with the definition of article 81, the equation of parabola in polar co- ordinates can be easily derived. In Fig. 75, let be the fixed point (focus), and D'D the fixed line (direc- trix). Choose as pole and OX, perpendicular to D'D, as the polar axis. Let P(p, 6) be any point on the locus. Draw MP and NP perpen- dicular to OX and D'D respectively. By definition, OP = NP. But OP = p, and NP = QM - p + p cos 6. Hence p = p + p cos 0. Solving for p, This is the polar equation of a parabola referred to its focus and axis. EXERCISES **X Fio. 75. 1. Given the equation p -, transform it to rectangular 1 - cos $ coordinates and by translation of axes derive the equation y* — 2px. 2. By taking the focus at the left of the directrix, derive the equation of the parabola in the form p = - — ■ • 1 + COS0 3. Change the following equation into polar coordinates with the Digitized by Google 112 ANALYTIC GEOMETRY [§92 pole at the origin, and the polar axis on the positive part of the x-axis: y* - 2px + p*. 4. Show that if the vertex of the parabola is taken as pole and the axis of the parabola as polar axis, the equation of the parabola in polar a : 2p cos coordinates is p = — r- = — • » sin 2 B 92. Construction of a parabola. — First method. — The directrix D'D and the focus F are supposed known. Place a right triangle, Fig. 76, with one side CB on the directrix as shown. Fasten one end of a string whose length is CA, at the focus F and the other at A. With a pencil at P, keep the string taut and move the triangle along the directrix, and the point P will generate a parabola, MT Fio. 76. Then FP = CP, Why? Second method. — As before, the directrix D'D and the focus F are supposed known. In Fig. 77, draw MX through F and perpendicular to D'D. Draw any number of lines A' A, B'B, etc., par- allel to the directrix, and intersecting MX in Mi, M 2f etc. With F as center and a radius equal to MMi, strike arcs intersecting A' A in Pi and Qi. In like manner, with MM 2 as a radius, strike arcs intersecting B'B. Continue in like manner for the other lines drawn. Then the points, thus determined, lie on the parabola. Why? In this way the parabola can be located as accurately as desired. D A B C *P> F Pi MiP M % M* ^4 Mt Pi P» « A' B' c' . & Fio. 77. Digitized by Google §93] THE PARABOLA AND CERTAIN FORMS 11 3» EXERCISES 1. Construct a parabola by. the second method, in which p = 1 in. In which p = $ in. 2. Construct a circle of radius 8 in., and a parabola with its vertex at the center of the circle, and its focus on the positive x-axis at the point midway between the center and circumference. Write the equa- tion of each in the standard forms, and compute the coordi- , nates of the points of inter- section of the curves. 3. Explain how the con- struction shown in Fig. 78, determines a parabola. APPLICATIONS 93. Parabolic arch. — The cable of a suspension p^ 78 bridge hangs in the form of an inverted parabolic arch. Arches for bridges, when the weight is uniformly distributed, are properly constructed in the form of a parabola. In metal-arch bridges the loading is practically uniform on the horizontal, and so such bridge structures are in the form of parabolic arches. The arches of concrete bridges jjM are seldom if ever built in the form of a parabola, for, in such structures, the load- ing cannot be uni- formly distributed on the horizontal. In the parabolic arch, Fig. 79, AB = 2s is the span, and CO = h is the height. If the origin is taken at the vertex of the parabola, and the axis along the y-axis, the equation is of the form x 2 = 2py. To find the value of p, we know that the point B(s, —A) is Digitized by Google 114 ANALYTIC GEOMETRY [§94 on the parabola. Substituting these coordinates in x 2 = 2py, s 2 gives s 2 = —2ph, and p = — st* Hence the equation of s 2 hx 2 the parabola is x 2 = — j-y, and from this y = 2 * The height of the arch at any distance x from the center is NP = NM + MP = h + y - A - ~ EXERCISES 1. A parabolic arch has a span of 120 ft. and a height of 25 ft. Derive the equation of the parabola, and compute the heights of the arch at points 10 ft., 20 ft., and 40 ft. from the center. 2. A parabolic arch has a span of 40 ft. and a height of 15 ft. Find the height of the arch at intervals of 5 ft. from the center. 3. The distance between the supports on the river span of the Brooklyn suspension bridge is about 1600 ft., and the vertex of the curve of the cables is 140 ft. below the suspension points. Find the equation of the curve if the lowest point is taken as origin. 4. The towers supporting a suspension bridge are 320 ft. apart and rise 80 ft. above the roadbed. The lowest point of the parabola formed by the cables is 20 ft. above the roadbed. Find the equation of the curve of the cables using as origin the point in the roadbed below the vertex of the parabola. 94. The path of a projectile. — A projectile starting at the origin, Fig. 80, with an initial velocity of v ft. per second, and making an angle a with \ the horizontal, would after t \ * seconds have the position x B = v cos at and y = v sin at, if the action of gravity and the resistance of the air were not considered. If the action of gravity is considered, y is decreased by \gt 2 ft. in t seconds. Then the coordinates of the projectile at time t are x = v cos a % and y = v sin at — \gt 2 . (I) Fig. 80. Digitized by Google §94] THE PARABOLA AND CERTAIN FORMS 115 The equation of the path of the projectile in rectangular coordinates is found by eliminating t between these equations and is ^"^^-srib^* 1 - (II) - EXERCISES 1. Eliminate t between the equations (I) and derive equation (II). 2. Show that equation (II) is a parabola with its vertex at the point c- 2 sin 2a v 2 sin 2 a\ , v 2 cos 2 < and p= — 2(7 2g ) y g 3. Find the s-intercept of (II), and thus find the range on the horizontal A . v 2 sin 2a to be 9 4. Find the height of the projectile when at a horizontal distance equal to one-fourth the range. 5. Find the horizontal range when v = 2000 ft. per second and (1) a - 45°, (2) a • - 30°, (3) a = 60°. Use g = 32. 6. Show that a projectile with a given velocity and at an angle of 60°, rises three times as high as it would if the angle were 30°. 7. What must be the initial velocity v of a projectile, if with an angle of elevation of 20°, it is to strike an object 80 ft. above the horizon- tal plane of the starting point, and at a horizontal distance of 1000 yd. ? GENERAL EXERCISES 1. The formula for the height of a bullet shot vertically upward with a velocity of 2000 ft. per second is s = 2000J - 16* 2 . Find the coordinates of the vertex, and plot the curve from which the height s at any time t may be read. 2. When one variable varies directly as the square of another, the equation connecting the two variables will represent a parabola. The length of a pendulum varies as the square of the time of a beat. This IT 2 gives the formula t 2 = -J, where t is time in seconds, g is 32, and I is y length in feet. Plot a curve from which can be read the time of a beat for lengths up to 20 ft. 3. In a parabolic reflector, such as used for an automobile headlight, the source of light is placed at the focus of the parabola that is a section of the reflector. Find the position of the source of light in a reflector 10 in. in diameter and 5 in. deep. Digitized by Google 116 ANALYTIC GEOMETRY [§94 4. Find the coordinates of the points of intersection of the parabola x* - Sy and the line 3s — 2y - 8 = 0. 5. Find the equation of the straight line passing through the focus of the parabola y* = &x and making an angle of 45° with the axis of the parabola. • 6. What value must be given to A; if the line 3x+2y + Aj = 0isto be tangent to the parabola x* =* — 6y? Plot. Suggestion. — Eliminate y between the two equations. Since a tangent meets the curve in two coincident points, the two values of x in the resulting equation must be equal. Hence put the discriminant of this quadratic equation equal to zero and solve for values of k. 7. Find the points of intersection of the following curves: x — Zy = and y* - 3s - Qy + 14 = 0. 8. For what values of m is the straight line y = mx + 2 tangent to the parabola x* - 6x + Sy + 41 = 0? 9. One end of a chord through the focus of a parabola is at the point (10, 10). Find the coordinates of the other end if the parabola has its vertex at the origin and its axis on the positive part of the rr-axis. 10. Transform the following equations in polar coordinates into rectangular coordinates and simplify: (1) " = 1 + cos 6 (2) " = 5 - 5 cos (3) " " nT'coTtf" 11. Plot the following curves given in polar coordinates and find the coordinates of their points of intersection: (1) p cos B - 4, p = -. — — -• (2) p - 4, p = 1 - cos v ' * ' H 1 + cos 12. Show that the equation p = 8 sec* $0 is that of a parabola, and sketch the curve. 13. Find the equation of the circle circumscribing the segment of the parabola y* = 2px, cut off by the latus rectum. 14. An equilateral triangle having one vertex at the origin is inscribed in the parabola y* — 2px. Find the length of a side of the triangle. 15. Show that x* + V* = a* is the equation of a parabola. Sketch the curve. 16. Find the equation of the parabola with x + y ™ as directrix, and focus at (ia, ia). Express in the form given in the previous exercise. Digitized by Google CHAPTER VII THE ELLIPSE AND CERTAIN FORMS OF THE SECOND DEGREE EQUATION . 95. The equation of the ellipse. — By the definition of article 81, the ellipse is the locus of a point whose distance from a fixed point, the focus, is to its distance from a fixed straight line, the directrix, in a constant ratio e, less than 1. In Fig. 81, let F be the focus and D'D the directrix. Choose as x-axis the line X'X through F and perpendicular to D'D at R. Fig. 81. Since e<l, there are two points V and V on X'X such that VF FV ~y = e and ^yr = e « Hence the points V and V are on the locus. Choose 0, the point midway between V and F', as origin, and Y'Y through 0, parallel to D'D, as y-axis. Let the length of W = 2a. Then VO = OV - a. 117 Digitized by Google 118 ANALYTIC GEOMETRY [§95 It is necessary first to find the equation of the directrix and the coordinates of the focus. From the definition of the ellipse, VF = e RV, or a - FO - e(RO - a), (1)' and FV = eRV\ or a + FO = e(RO + a). (2) Adding equations (1) and (2), 2a = 2eR0, or 720 = -• e , . Then the equation of the directrix is x = — • r Subtracting equation (1) from equation (2), 2F0 = 2ae, or FO = ae. Then the coordinates of the focus F are ( — ae, 0). Now to derive the equation, let P(z, y) be any point on the locus* and draw FP, and NP perpendicular to D'D. By definition, FP = ei\TP. But FP - Vte + ae) 2 + y 2 , and iVP - ? + *. Then V(x + ae) 2 + y* = e(^ + x) . Squaring and arranging, this becomes a 2 ^ a 2 (l - e 2 ) ' Since e < 1, a 2 (l — e 2 ) is positive and less than a 2 . Let it be represented by b 2 and the equation of the ellipse is x 2 v 2 This is a standard form of the equation of the ellipse, and is . the form in which the equation of the ellipse is usually written. Its simple form is due to the choice of the coordinate axes. A different choice of axes would give a less simple form of the equation, but the locus itself would be unaltered. Va 2 — b 2 Since b 2 - a 2 (l - e 2 ), e - -^ — 7 a Digitized by VjOOQIC §96] THE ELLIPSE AND CERTAIN FORMS 119 . Equation [32] is the required equation of the ellipse. For it has been proved true for every point on the ellipse, and it can be readily proved that it is not true for any point that is not on the locus. The proof of this is left as an exercise. 96. Shape of the Ellipse. — The shape of the ellipse and its position relative to the coordinate axes can be readily x 2 y 2 determined from the equation — 2 + ^ = 1. Solving for x, x = iiW-?- Solving for.y, y = ± - Va 2 — x 2 . (1) For all values of y such that b 2 — y % > 0, x has two real values, numerically equal but opposite in sign. Wheny 2 = b 2 , x = 0. For all values of x such that a 2 — x 2 > p, y has two real values, numerically equal but opposite in sign. When x 2 = a 2 ,y = 0. Hence the curve is symmetrical with respect to both coordinate axes and the origin, and its intercepts are a and —a on the rc-axis, and b and —6 on the y-axis. (2) For all values of y such that b 2 — y 2 <0, x is imaginary; and for all values of x such y that a 2 — x 2 <0, y is im- aginary. Hence no part of the curve lies outside of the rectangle bounded by the four lines x = ±a and y = ±6. (3) As x increases from —a to 0, the positive value of y increases from to 6, and the negative value of y decreases from to —6. As x increases from to a, the positive value of y decreases from b to 0, and the negative value of y increases from —6 to 0. The ellipse has the shape shown in Fig. 82. CO. 6) B V-6 1 II a V ^ 9 It H v (-«.0) F O (a,0 (0.-6) B' »=-& Digitized by Google 120 ANALYTIC GEOMETRY .[§97 The formula ft 2 = a 2 (l — e 2 ) can now be readily interpreted geometrically. For in the right triangle FOB, Fig. 82, FO= ae and OB = 6. ___ Then FB 2 = (ae) 2 +6 2 . But from ¥ - a 2 (l - e 2 ), a 2 - (ae) 2 + 6 2 . Hence a 2 = FB 2 , or a = FB. >-x 97. Definitions. — The center of symmetry of the ellipse is called the center of the ellipse. The chord through the focus and center of an ellipse is called the major axis. Its length is 2a. One-half of the major axis is called the semimajor axis. The chord through the center of the ellipse and perpendicular to the major axis is called the minor axis. Its length is 26. One-half of the minor axis is called the semiminor axis. The chord of the ellipse through the focus and perpendicular 26 2 to the major axis is called the latus rectum. Its length is — t for the abscissa of the focus is — ae, and when x = — ae, The points on the ellipse at the ends of the major axis are the vertices of the ellipse. In Fig. 83, V'V is the major axis, B'B the minor axis, and P'P the latus rectum. An ellipse can be readily sketched if the position and lengths of the axes are known. 98. Second focus and second directrix. Theorem. — An ellipse has two foci and two directrices. In Fig. 84, on OV take OF' = FO and OR' = RO. Draw E'E parallel to D'D. Then F' is also a focus and E'E the corresponding directrix of the ellipse. Digitized by Google §99] THE ELLIPSE AND CERTAIN FORMS 121 Proof. — Let P be any point of the ellipse. Thrbugh P draw PN parallel to the rc-axis and intersecting D'D in N. Because of the symmetry of the ellipse, PN intersects the ellipse at a second point P' and the line E'E at N'. Draw PF and P'F'. From the symmetry of the figure, FP « F'P', and NP = P'N'. F'P' FP But NP = € ' P'N' = e. Then the ellipse is also the locus of a point P' whose distance from F' divided by its distance from E'E is e. Therefore F' is a focus and E'E is the corresponding directrix of the ellipse, and the ellipse has two foci and two directrices. The coordinates of the foci are (±oe, 0), and the equations of the directrices are x = +— Fig. 84. Fig. 85. 99. Ellipse with major axis on the y-axis. — The equation of an ellipse whose major axis is on the t/-axis, and whose center is at the origin is obviously obtained by interchanging x and y in the work of article. 95. The equation then is V 2 X 2 [33] ?i + £-,-l. Here the major axis is 2a as before; the minor axis is 26; the coordinates of the vertices are (0, ±a); the coordinates of Digitized by Google 122 ANALYTIC GEOMETRY [§99 the foci are (0, ±ae); and the equations of the directrices are EXERCISES 1. In each of the following ellipses find the semimajor axis, the semi- minor axis, the eccentricity, the coordinates of the foci, and the equa- tions of the directrices. Sketch each ellipse. (3) 4x* + 9y* - 36. (6) 6s* + 9y* - 54. 2. Find the distance from the foci to the ends of the minor axis in x 1 y 1 the ellipse — „ + tz = 1. a 2 o z 3. Write the equation of an ellipse with center at the origin, and major axis on the x-axis, having given: (1) o - 6, b - 4. (4) Focus at (5, 0), e - J. (2) a = 4, c = J\/3. (5) Directrix is x =7, e = }. (3) 6 = 3, e = J. (6) Latus rectum = 4, a = 8. (7) Focus at (\/3, 0), directrix is x = 3-\/3. £ 2 v 2 4. In the ellipse — + — = 1, find the values of y when x = 2, when 25 16 x «■ 4, when £ = 5, when x = 6. x 2 y* 5. Find the length of the latus rectum in the ellipse — + r- «■ 1. x* y* In the ellipse — + — = 1. 6. Find the equation of the ellipse with center at the origin, axes on the coordinate axes, and passing through the points (1, f\/3) and (K/6, 1). 7. Derive equation [33] from [32] by rotating the coordinate axes through an angle <p = 90°. 8. Find the semi-axes, eccentricity, and the latus rectum of each of the following ellipses: (1) 6y* - 30 - 5s*. (2) 2x* + y* - 2m, ro>0. (3) x* + qy* - «, g>l, and s>0. (4) px* + gy* = pg, p>0, g>0, and g>p. Digitized by Google §100] THE ELLIPSE AND CERTAIN FORMS 123 9. Find the distances from the foci of the ellipse — + •— = 1 to a point on the ellipse, whose abscissa is 2. 10. The minor axis of an ellipse is 24, and the foci and origin divide the major axis into four equal parts. Find the equation of the ellipse. 11. Assume the equation of the ellipse, — + r. = h BJi ^ show that the sum of the distances of any point on it from its foci is 2a. 12. Regard the circle as an ellipse with a — b f and find its foci, direc- trices, and eccentricity. 13. Find the coordinates of the points of intersection of the ellipse 2x* + Zy* = 14 and the parabola y* — 4x. 14. Find the locus of the vertex of a triangle if the base is 2a, and b* the product of the tangents of the angles at the base is — c* Suggestion. — Take the x-axis on the base and the origin at the center. 15. Find the locus of the vertex of a triangle if its base is 26 and the sum of the other sides is 2a. Take the x-axis on the base and the origin at the midpoint. x^ t/' x' t/* 16. Discuss the equations — -f ti = 0, and — + — = —1. The first a 2 o* a* o* of these is the equation of a point ellipse and the second is that of an imaginary ellipse. 100. Equation of ellipse when axes are translated. — Transform the equation — + — = 1 by translating the coor- dinate axes to a new origin at a point 0'(—h, — k), using [12], and we have (x' - A) 2 + ^^- 2 = l. This is the equation of an ellipse having its center at the point (h, k) referred to the new coordinate axes, and having its axes parallel re- spectively to the rc'-axis and the j/-axis, as shown in Fig. 86. F' o' (-/».-*) Fig. 86. If the primes are dropped, this equation becomes ->* +x' Digitized by Google 124 ANALYTIC GEOMETRY [§101 M fe^C + fc^-,, which is a second standard form of the equation of the ellipse, and is a convenient form for writing the equation of an ellipse with center at the point (A, k) and major axis parallel to the x-axis. Similarly, the equation of an ellipse with center at (A, k) and major axis parallel to the y-axis is of the form 1*1 fc^SI + fe^E-i. Example. — Find the equation of an ellipse with semimajor axis 5, semiminor axis 4, center at point (3, —2), and major axis parallel to the o;-axis. Substituting in [84], {x ~ 3)> + (y + 2)> - 1. Y Simplifying, 16s* + 25y* - 96* + lOOy - 156 - 0. The ellipse is shown in Fig. 87. 7 EXERCISES 1. Write the equations of the fol- lowing ellipses, and plot: (1) Center at (3, 4), a = 5, b = 3, and major axis parallel to x-axis. Y (2) Center at (-3, -7), a - 6, Fio. 87. & = iV% and major axis parallel to jfcaxis. 2. Find the equations of the two ellipses, each having its center at (5, — 4), o = 6, and 6=4; one having its major axis parallel to the z-axis and the other having its major axis parallel to the y-axis. 3. Find the coordinates of the foci and the equations of the directrices of each ellipse of exercise 1. 4. Find the equation of the ellipse with center at (10, 2), one directrix the line x = 2, and eccentricity f . 5. Find the equation of the ellipse with center at (3, 4), major axis parallel to x-axis, and passing through the points (—2, 4) and (3, 0). 6. Find the equation of the ellipse having its center at (4, 2), major axis parallel to the y-axis, semimajor axis 6, and passing through the point (8, 4). Digitized by Google §101] THE ELLIPSE AND CERTAIN FORMS 125 101. Equation of the form Ax 2 + Cy 2 + Dx + Ey + F =0. Every equation of the form Ax 2 + Cy 2 + Dx + Ey + F = 0, where A and C have like signs but different values, represents an ellipse with axes 'parallel to the coordinate axes. Proof— Given Ax 2 + Cy 2 + Dx + Ey + F = 0. Completing the squares in x and in y, D\ 2 , nl , E\ 2 CD 2 + AE*-4ACF A i* + U) + C (" + £) " 4AC Dividing by the second member of this equation, (*+2z) 2 ( y+ &Y t CD 2 + AE 2 - 4ACF + CD 2 + AE 2 - 4ACF " lm AA 2 C 4AC 2 This is in the form of [34] if A<C where D . tf t CD 2 + AS 2 - 4ACF A = ~2A' * = ~2C' a = il^C ' and CD 2 + AE 2 - 4ACF 6 " IAC 2 and therefore represents an ellipse with axes parallel to the coordinate axes if A and C have like signs so that 44 2 C and AA C 2 have like signs. It is of the form of [34J if A <C. From the preceding, it follows that the equation of an ellipse in the form Ax 2 + Cy 2 + Dx + Ey + F = can be trans- formed into one of the forms [32] or [33] by a suitable trans- lation of the coordinate axes, the new origin being at the point (~^-^)- Example 1. — Transform to the second standard form, the equation of the ellipse 24a* + 49y* - 96s + 294y - 639 = 0, find the coordinates of the center, foci, and vertices, the length of the semimajor and semi- minor axes, and the equations of its directrices. Plot. SolvJUon. — Completing the squares in x and in y, 24(s* - 4* + 4) + 49(y* + Gy + 9) = 639 + 96 + 441, or 24(z - 2)* + 49(y + 3)* = 1176. Dividing by 1176 and putting in the form of [84], (s - 2)* (y 4- 3)' - 49 T 24 Digitized by Google 126 ANALYTIC GEOMETRY [§101 +*.x This is an ellipse (Fig. 88) with center at the point C(2, —3) and axes parallel to the coordinate axes. The semi axes are a — 7, and b = 2\/S. The eccentricity e = ^° — — = =• a 7 The distance from the center to the foci is ae = 5, and the foci are >(7, -3)and*"(-3, -3). The vertices are 7(9, -3) and 7'(-5, -3). The distance from the center to the directrices is - = fy and the e equations of the directrices are x — y and x = — V- The ellipse is as shown in the figure. • Example 2. — Find the equation of the ellipse whose axes are parallel to the coordinate axes and which passes through the points (—2, 7), (2, 4), (-2,1), and (-6,4). Solution. — The required equation is of the form Ax* + Cy* + Dx + Ey +F = 0. If this is divided by A the equation is of the form x s + C'y* + Jyx + E'y + F' - 0, and therefore contains only four arbitrary constants, which can be found from four equations. Dropping the primes and substituting the coordinates of the four given points, 4 + 49C - 2D + 7E + F - 0, 4 + 16C + 2D + 4JS + F = 0, 4 + C-2D+E+F = 0, 36 + 16C - 6D + \E + /? = 0. Solving, C-V»I>-4,J£» -H*' ^ = H 1 - The required equation is x* + W + 4* - *Vy + H* = o, or 9z 2 + 16y 2 + 36* - 128y + 148 = 0. Example 3. — Translate the coordinate axes so that the equation of the ellipse 4s 2 + 9y 2 - 24s - 36y + 36 = is in the form [32]. Solution. — Completing the squares in x and in y f 4(s 2 - 6s + 9) + 9(t/ 2 - 4y + 4) = -36 + 36 + 36. Whence fe^ + <L^»! _ i. Digitized by Google §102] THE ELLIPSE AND CERTAIN FORMS 127 Putting a; — 3»s / , or x = x 7 + 3 and y — 2 -» y 7 , or y = y* + 2, 9 -V2 **X Fig. 89. This is of the form [32], and is an ellipse referred to coordinate axes that are parallel to the old coordinate axes, and with the new origin at the point (3, 2). The ellipse is as shown in Fig. 89. The transformation could evi- dently be made by substituting x = x 1 + h and y = y f + k } and- proceeding as in the example of article 89. EXERCISES 1. Express the equations of the following ellipses in the form [34] or [34i], find the coordinates of the centers, foci, and vertices, the lengths of the semimajor and semiminor axes, and the equations of the directrices. Plot each. (1) 7z* + 16y* + 14a? - 64t/ - 41 - 0. (2) 8x* + 4y 2 - 64^- Sy + 68 - 0. (3) 4z* + 9y 2 - 8s + ISy + 12 - 0. (4) 8s* + 9y 2 + 16* - 54t/ - 1 = 0. 2. Transform 6s 2 + 7y 2 — 36s + 14y -f 53 = to new axes parallel respectively to the old axes, with the new origin at (3, —1). 3. Transform each of the ellipses of exercise 1 to the form [32] or [33],. find the coordinates of the foci, and the equations of the directrices referred to the new coordinate axes. 4. Find the equation of the ellipse with major axis parallel to the a5-axis, and center at the point (—3, 4), eccentricity $» and passing through the point (6, 9). 5. Find the equation of the ellipse with one focus at the point (6, 2), corresponding directrix the line x = 12, and eccentricity i- 6. Transform the following equation to one in which there are no x and y terms, and plot: 9s 2 + 12y 2 — 18z — 72y + 9 = 0. 7. Find the equation of the ellipse with eccentricity i, a focus at the point (2, 0), and the corresponding directrix the line x + 2 ■» 0. 102. Equation of ellipse when axes are rotated. — In article 90 it was seen that when the coordinate axes were rotated Digitized by Google 128 ANALYTIC GEOMETRY [§102 through an angle <p, a term in xy appeared in the equation of the parabola. Likewise if the equation of an ellipse with axes parallel to the coordinate axes, Ax 2 + Cy 2 + Dx + Ey + F = 0, is transformed by using formulas [13], the equation takes the form Ax* + Bxy + Cy 2 + Dx + Ey + F = 0. This is the most general form of an equation of the second degree in x and y, where B* - 4AC<0. (See Art 122.) Conversely, starting with an equation containing an xy- term, rotation through a properly chosen angle will cause the xy-term to disappear by having its coeffi- cient zero. Example 1. — Transform the equation 9x* + 16y 2 - 36a; - 96y + 36 = 0, by ro- tating the coordinate axes through an angle of 30°. Sketch the ellipse. Solution. — Using formulas [13], x = x' cos 30° - if sin 30° - ly/S* - W - i( a/3*' - V'), and y = x> sin 30° + \f cos 30° = \x' + \y/%\f - W + y/ltf). Substituting these values in the given equation and simplifying, 43« , *+14v^a;yH-57y , *-24(3VS+8)x , -24(8V3--3)y , -|- 144 = 0. The ellipse and the two sets of coordinate axes are sketched in Fig. 90 EXERCISES 1. Transform the following equations by rotating the coordinate axes through the angle given in each case: f+-X Fig. 90. wS+B-i- <P = 45° (2) 16s* + 9y* - 144. <p - 60°. (3) 3ftr* + 4y» = 144. <p = 90°. (4) 2x* + 3y* - 4* + Zy - 10 = 0. *> - 30°. (5) s* + xy + y 2 + 2x + y + 2 = 0. *> = 45°. (6) 6s* + 4sy + 6y* + 5s - Sy = 0. *> = tan" 1 J. 2. Transform the following equation to the standard form by rotating the axes: 29s 2 + 16sy + 41y* - 45 = 0. Sketch the ellipse with both sets of axes. 3. Simplify the following equation by first translating the axes to remove the z-term and the y-term, then by rotating through an angle Digitized by Google §103] THE ELLIPSE AND CERTAIN FORMS 129 that will remove the xy-term. Sketch the curve and the three sets of coordinate axes: x* + xy + y* + 2x + 3y — 3 =0. 103. Equation of ellipse in polar coordinates. — In a manner similar to that of article 91, the equation of the ellipse in polar coordinates may be derived. EXERCISES 1. Derive the equation of an ellipse with the pole at the focus to the right of its corresponding directrix, and the polar axis perpendicular to the directrix. Also derive the equation when the focus is taken at the left of its corresponding directrix. Let p equal the distance from the focus to the directrix. 2. Transform the results of exercise 1 to rectangular coordinates, and change to the standard form by translation of axes. 3. Derive the polar equation of an ellipse, the pole being at a focus, Cx — c\ ' i/' by starting with the equation 5 \- p- = 1, and then putting x = p cos 0, y = p sin 0, c — ae, and b* — (1 — e 2 )o 2 ; finally solving the quadratic equation for p. 4. Derive the polar equation of an ellipse, the pole being at the center and the polar axis along the major axis. 5. Show that pe in exercise 1 is one-half the latus rectum* 104. Construction of an ellipse. — First method. — The length of the major axis 2a and the foci F and F f are supposed to be known. On a drawing board fasten the ends of a string of length 2a at F and F', Fig. 91. Place a pencil point, P, in the string and move it about keeping the string taut. Then the point P will generate an ellipse. This construction depends upon the following: Theorem. — The sum of the distances from any point on an ellipse to its foci is constant and equal to the major axis. < This may be proved as follows: In Fig. 92, from the definition of an ellipse, Fig. 91. Digitized by Google 130 ANALYTIC GEOMETRY [§104 and PF' - e-N'P = e(^ + x) = a + ex, PF = e-PN = eg-z) = a- err. Adding, PF' + PF = 2a = major axis. Second method. — The major axis, 2a, and the minor axis, 26, are supposed to be known, as well as the position of the center and direction of axes. With the center of the ellipse as a center describe two circles of radii a and 6 respectively, Fig. 93. Draw any radius intersecting the inner circle in R and the outer circle in Q. Through R draw a line parallel to the major axis, and through Q a line parallel to the minor axis. Then the point P where these lines intersect is a point on the ellipse. In this manner any number of points on the ellipse can be determined. That the point P is on the ellipse with its major axis on the x-axis can be proved as follows: N D i J \f' o F J € a D' - ~ < - > •*x Fig. 92. Fig. 93. y z Equation of ellipse is — 2 + ~ = 1. a o If is the angle XOQ, the coordinates of P are x = OM = a cos B y and y = MP = b sin 6, Substituting in the equation of the ellipse, a 2 cos 2 6 , 6 2 sin 2 6 n , . — ~2 h r^ — = cos 2 6 + sin 2 = 1. Digitized by Google §105] THE ELLIPSE AND CERTAIN FORMS 131 Hence the equation is satisfied and the point P is on the ellipse. EXERCISES 1. By the second method construct an ellipse having semiminor and semimajor axes 1 in. and 1J in . respectively. 2. Prove that the projection of a circle upon a plane making an acute angle with the plane of the circle is an ellipse. APPLICATIONS 105. Uses of the ellipse. — The ellipse is involved in many practical considerations, as well as being frequently used in mathematics and its applications. It was believed by tfie ancient Greeks that the sun was the center of the universe in which we live. Kepler (1571-1630) stated that the orbits of the planets are ellipses. Newton (1642-1727) showed that the law of gravitation determines the orbits to be ellipses. In architecture, because of the beauty of its form, the elliptic arch is frequently used. Some noted structures were built in the form of an ellipse. The Colosseum at Rome was of this form. In bridge structures, many of the most noted stone-arch bridges of the world are elliptical. In machinery, elliptical gears are often used where change- able rates of motion are desired, as in shapers, planers, and slotters where the cutting speed is less than the return motion. In the study of electricity and mechanics, the ellipse is frequently used. EXERCISES 1. The Colosseum at Rome is in the form of an ellipse 615 ft. long and 510 ft. wide. Find the equation of the ellipse and the position of the foci. 2. A stone-arch of a bridge has a span of 200 ft. and a height of 42 ft. The arch is in the form of a semi-ellipse. Find the equation of the ellipse and the position of the foci. Digitized by Google 132 ANALYTIC GEOMETRY [§105 3. In exercise 2, find the heights of points 50 ft. and 25 ft. from one end of the arch. 4. In considering equipotential surfaces in electricity, the equation . , x + — ; x = 1 is used. If a > b and X denotes an arbitrary con- a* + A o* + A slant, such that X > — b* show that the equation represents a system of ellipses haying the same foci. 5. An arch is in the form of a semi-ellipse with major axis horizontal. The span is 80 ft. and the height is 30 ft. Find the distance of the arch below the level of its top for each 10 ft. of the span. 6. The earth's orbit is an ellipse with the sun at one focus. The major axis is 185.8 million miles and the eccentricity is about A* Find the difference between the greatest and the least distance from the earth to the sun. 7. Show that, if two equal elliptical gears turn on mountings at corresponding foci, they are always in contact. 8. If two equal elliptical gears have major axes and minor axes of 12 in. and 8 in. respectively, and revolve once in 10 seconds, find the greatest and the least linear speed of a point on the driving ellipse. Suggestion. — Use the greatest and the least radius on which a point is turning. The driving gear has uniform angular velocity, and the mountings are at corresponding foci. '• GENERAL EXERCISES 1. Find the equation of an ellipse in the form of [32] having the sum of its axes 20, and the difference 4. 2. Find the equation of an ellipse in the form of [33] if its major axis is 24, and its minor axis is equal to the distance between the foci. 3. Find the equation of an ellipse in the form of [32] if the minor axis is 12, and the distance between the foci is 12. 4. Find the equation of the ellipse in the form of [33] in which o = 8, and the foci bisect the semi major axes. Find the semi-axes, coordinates of foci, eccentricity, and the equa- tions of the directrices of each of the following ellipses: 5. 16x 8 + 9y 8 = 144. 6. 24x 8 + 36y 8 = 864. 7. 16s 8 + 25y 2 - 64a; + lOOy - 236. Transform each of the following equations to axes parallel respec- tively to the old, the new origin being at the point given in each case. Plot the curve and both sets of axes. Digitized by Google §105] THE ELLIPSE AND CERTAIN FORMS 133 8. 9x* + 4y* + 3ftr - 24y + 36 - 0. (-2, 3). 9. 25s 8 + 16V + 50x + 32y - 359 - 0. (-1, -1). 10. Derive an equation that will represent all ellipses having foci at the points (3, 0) and (-3, 0). 11. Derive the equation of the ellipse with a focus at (3, 1), eccentricity equal to }, and with Zx — 4y + 6 «■ as directrix. 12. Show that the latus rectum of an ellipse is a third proportional to x s y 1 the two axes. Find the latus rectum of the ellipse — + — : — 1 by this method. Digitized by Google CHAPTER VIII THE HYPERBOLA AND CERTAIN FORMS OF THE SECOND DEGREE EQUATION 106. The equation of the hyperbola. — By the definition of article 81, the hyperbola is the locus of a point whose distance from a fixed point, the focus, is to its distance from a fixed straight line, the directrix, in a constant ratio e, greater than 1. The method used in deriving the equation is exactly the same as that for the ellipse, Art 95. In Fig. 94, let F be the focus and D'D the directrix. Choose as x-axis the line X'X through F and perpendicular to D'D at R. Since e > 1 there are two points V and V on X'X such VF V'F that ^y = e and y^ = e. Hence the points V and V are on the locus. Choose 0, the point midway between V and V' y as origin, and Y'Y f parallel to D'D, as t/-axis. Let the length of V'V = 2a. Then V'O = OV = a. 134 Digitized by Google §106] THE HYPERBOLA AND CERTAIN FORMS 135 As with the ellipse it is necessary to find the equation of the directrix and the coordinates of the focus. From the definition of the hyperbola, VF = e RV, or OF - a = e(a - OR), (1) and VF = eV'R, or OF + a = e(a + OR). (2) Subtracting equation (1) from equation (2), 2a = 2eOR, or OR = -• Then the equation of the directrix is x = — Adding equations (1) and (2), 20F = 2oe, or OF = ae. Then the coordinates of F are (ae, 0). To derive the equation of the hyperbola, let P(x, y) be any point on the locus, join F and P, and* draw iVP perpendicular to D'D. By definition, FP = e i\TP. But FP ** V(* - <*0 2 + 2/ 2 , and NP = x - j. Then V(s - a*) 2 + y 2 = e(s T fj • Squaring and arranging, this equation becomes x 2 y 2 a 2 ~ a 2 (e 2 - 1) = Since e>l, a 2 (e 2 — 1) is positive. Let it be represented by b 2 and the equation of the hyperbola is x 2 v 2 This is a standard form of the equation of the hyperbola, and is the form in which the hyperbola is usually written. Its simple form is due to the choice of the coordinate axes. A different choice of axes would give a less simple form of the equation, but the locus would be unaltered. Since b 2 = a 2 (e 2 -!),« = ^ a% + b *- Digitized by Google 136 ANALYTIC GEOMETRY [5107 Equation [35] is the required equation for it has been proved true for every point on the hyperbola, and it can be readily proved that it is not true for any point that is not on the locus. 107. Shape of the hyperbola. — The shape of the hyperbola and its position relative to the coordinate axes can be readily x 2 t/ 2 determined from the equation — 2 — 4- t = 1. Solving for x, Solving for y, (1) For all values of y, x has two real values, numerically equal but opposite in sign. For all values of x such that x - ± j^Vbf+V*. y hs ± -V x* — a 2 '(-a.O) o (a.0)l (<w.O) +x' Fig. 95. x 2 — a 2 >0, y has two real values, numerically equal but opposite in sign. When x 2 = a 2 , y = 0. Hence the curve is symmetrical with respect to both coordinate axes and the origin, and its intercepts on the x-axis are a and —a. (2) For all values of x such that x % — a 2 <0, y is imaginary, but no value of y will make x imaginary. (3) As x increases from + a or decreases from — a, the positive values of y increase and the negative values of y decrease. The hyperbola has the shape shown in Fig 95. Digitized by Google §108] THE HYPERBOLA AND CERTAIN FORMS 137 108. Definitions. — The center of symmetry of the hyperbola is called the center of the hyperbola. The line through the focus and perpendicular to the directrix is called the principal axis of the hyperbola. The points in which the hyperbola intersects the principal axis are called the vertices of the hyperbola. The portion of the principal axis lying between the vertices is called the transverse axis of the hyperbola. Its length is 2a. The conjugate axis of the hyperbola has a length 26, is perpendicular to the principal axis, is bisected by it, and passes through the center. The chord of the hyperbola through the focus and perpen- dicular to the principal axis is called the latus rectum. Its 26* length is — > for the abscissa of the focus is ae, and when r E a x = ae,y 109. Second focus and second directrix. — The hy- perbola -j — p = 1 has a second focus at the point (— ae, 0), and a second direc- trix which is the line x = a e The proof is similar to that of article 98 for the ellipse and is left as an exercise for the student. In Fig. 96, F and F 9 are the foci, and the lines D'D and E'E are the directrices. 110. Hyperbola with transverse axis on the y-axis. — The equation of an hyperbola whose transverse axis is on the y-axis and whose center is at the origin is obtained by inter- Digitized by Google 138 ANALYTIC GEOMETRY [5110 changing x and y in the work of article 106. then y 2 x 2 The equation is [36] J -„ - ,-. = 1. Here the transverse axis is 2a; the conjugate axis is 26; the coordinates of the vertices are (0, ±a); the coordinates of the foci are (0, ±ae); and the equa- tions of the directrices are y = +-• 6 (See Fig. 97.) EXERCISES 1. In each of the following hyperbolas, find the length of the transverse axis and the conjugate axis, the coordinates of the foci, and the equations of the directrices. Sketch each hyperbola. (6) 9y* - 25s* - 225. 2. Write the equation of the hyperbola with center at the origin, and transverse axis on the s-axis, having given: (1) p * - 6, b - 4. (4) b = 3, oe = 5. (2) a = 4, e »» 2. (5) a - 9, e - $• (3) 6 = 8, ae - -%S/5- (6) 6 - 6, at - \/85. x* y* 3. In the hyperbola - — — = 1, find the value of y when x » 3, when x = 5, when 3 = 2. 4. Find the length of the latus rectum of the hyperbola — — — = 1. 36 lo y 2 x 2 Of the hyperbola — - — — = 1. a 2 b 2 5. Find the equation of an hyperbola with transverse axis on the x-axis, center at origin, and passing through the points (6, 4) and ( —3, 1). 6. Find the equation of the locus of a point moving so that the differ- ence of its distances from the points ( ±6, 0) is 8. Fig. 97. <■>§-?«->■ (2) x - - J?- . 1 W 36 100 *' (3) 16a; 1 - 9y l = 144. Digitized by Google §111] THE HYPERBOLA AND CERTAIN FORMS 139 7. Derive equation [36] from [35] by rotating the coordinate axes through an angle <p = 90°. 8. Find the semi-axes, eccentricity, and the latus rectum of each of the following hyperbolas: (1) ii- i. (4) - " V* - m. (2) 4x* - 3y* - 24. (5) px* - qy* « pq. (3) 16x* - y* - 16. (6) s a - qy* - «. 9. Find the semi-axes, coordinates of foci, eccentricity, and equations of directrices of each of the following hyperbolas: (1) 16s 8 - 9y* - 144. (2) 24x 8 - 36y 8 - 864. 10. Find the equation of an hyperbola with transverse axis on the y-Sixia, center at the origin, eccentricity equal to 2, and passing through the point (3, 2). x 8 y* 11. Assume the equation --- = 1 of the hyperbola, and show a 2 o 2 that the difference of the distances of any point on it from the foci is 2a. 12. Show that the latus rectum of an hyperbola is a third proportional to the two axes. 13. What does the equation x % — y % = 16 become when the coordinate axes are rotated through an angle <p = —45°? 14. Find the equation of an hyperbola if its center is at the origin, transverse axis is 24, and the distance between its foci is 32. 15. Find the equation of an hyper- bola if its center is at the origin, transverse axis is 24, and its con- jugate axis equals one-half the dis- tance between its foci. 111. Asymptotes. — In Fig. 98, P'P is a line passing through the* center of the hyperbola and in- tersecting the curve in P' and P. If P is made to move off to infinity along the curve, the line P'P \ continually passing through v A * the center, will turn about and will approach one of the two lines A 1 A or B'B. These lines are called the asymptotes 1 of x the hyperbola. 1 This is not the general definition for asymptotes, but is true for the hyperbola. +x Digitized by Google 140 ANALYTIC GEOMETRY [§111 The equation of any line P'P through the origin is y =* mx. The coordinates of its intersection with the hyperbola 6* 1 are found by solving the equations as simultaneous equations. Solving for x, x = ± ab Vb* - a*m 2 ' Now as P moves off to infinity along the curve x becomes infinite. Therefore the denominator of the fraction must approach 0. This gives 6 2 — a*m 2 = 0, or m = ± — Hence the equations of the asymptotes are [37] y - -x, and y -x. These equations can be combined into the single equation a* b* U * The conjugate axis B'B, Fig. 99, can now be brought into a closer relation to the hyperbola. If through the extremities of B'B lines are drawn parallel to the transverse axis, and through the ex- tremities of the transverse axis V'V lines are drawn parallel to the conjugate axis, a rectangle is formed with its diagonals on the asymptotes of the hyperbola. It can readily be shown that if the transverse axis of the hyperbola is on the #-axis, the equations of the asymptotes are y and V--5*. Digitized by Google §112] THE HYPERBOLA AND CERTAIN FORMS 141 By the help of the asymptotes, a simple and fairly accurate method for sketching an hyperbola is as follows: Locate the vertices and draw the asymptotes, then draw the hyperbola so that the curve continually approaches the asymptotes as it nfoves off toward infinity. Example. — Sketch the hyperbola 16x» - 25y* - 400. First put 16s* - 25y* - 400 in the * Then a = 5, b = 4, the foci are at the points (5, 0) and (—5, 0), and the equations of the asymptotes are y = ix and y - -£r. p^. 100# The curve is as shown in Fig. 100. 112. Conjugate hyperbolas. — Two hyperbolas that are so related that the transverse axis of each is the conjugate axis Fig. 101. of the other, both in magnitude and in position, are called conjugate hyperbolas. Thus, ^= — r; = 1 and r? — ^= =* 1, Fig. 101, are conjugate hyperbolas. Digitized by Google 142 ANALYTIC GEOMETRY [§113 From article* 111, it is seen that the asymptotes of each are y = ±-z. Therefore two conjugate hyperbolas have the same asymptotes. The formula 6 2 = a 2 (e 2 — 1) can now be readily interpreted geometrically. For in right triangle OVN, Fig. 101, OV = a, VN = OB = 6, and ON = OF = ae. 113. Equilateral hyperbola. — If a = 6, the hyperbola x 2 y 2 — 2 ;— t- " 1 becomes x 2 — j/ 2 = a 2 . This is called an equilateral hyperbola. The equations of its asymptotes are y = ±x, and are evidently perpendicular to each other and make angles of 45° with the axes of the hyperbola. An equilateral hyperbola is also called a rectangular hyperbola. It may be noted that the equilateral hyperbola is the simplest of hyperbolas, just as the circle is the simplest of ellipses, being the ellipse in which the major axis and minor axis are equal. EXERCISES 1. Find the equations of the asymptotes and sketch the curve for each of the following hyperbolas: (1) * _ £ ' . 1 ( 4 ) Ml » £* = 1 K } 18 12 l ' w 25 16 * (2) 9z* - 18y* = 16. (5) x* - y* - 12. (3) 9s 2 - 1%* = -16. (6) x* - y 2 - -12. 2. One of two conjugate hyperbolas is 12z 2 — I62/ 2 = 192, find the other. Find the coordinates of the foci and the equations of the direc- trices of each. 3. Show that the four foci of two conjugate hyperbolas, and the four points of intersection of the tangents at their vertices, all lie on a circle whose center is at the common center of the two hyperbolas. 4. Show that the eccentricity of an equilateral hyperbola is \/2. 6. Transform the equation of the equilateral hyperbola x* — y % — a*, by rotating the coordinate axes through an angle <p = —45°. This refers the hyperbola to its asymptotes as axes. Digitized by Google §114] THE HYPERBOLA AND CERTAIN FORMS 143 6. Find the equation of the hyperbola whose vertices are at ( ±4, 0) and the angle between whose asymptotes is 60°. 7. If ei and e a respectively. are the eccentricities of two conjugate hyperbolas, show that ae x « bet and that — = H — = -*» 1. ei z e** 8. Plot the equilateral hyperbolas x* — y % = a* and y* — x* = a* and locate their foci. With the same coordinate axes plot the circle x % + y 2 = 2a*. Also plot x 1 — y* = on the same set of axes. 9. Prove that in any hyperbola the distance from a focus to an asym- ptote equals the semi-conjugate axis. 10. Prove that in any hyperbola the distance from the center to the foot of the perpendicular from a focus to an asymptote equals the semi- transverse axis. 11. Find the value of b in order that the line y = 2x -f b shall be • x* y* tangent to the hyperbola = 1. 9 4 12. Find the value of m in order that the line y = mx -f 2 shall be x 1 y* tangent to the hyperbola — — — = 1. 16 9 114. Equation of hyperbola when axes are translated. — By a method identical to that of article 100 for the ellipse, the equation [38] — - 2 ^ 1 is found for the hyperbola with its center at the point (h, fc), and whose transverse axis is parallel to the x-axis. This is a second standard form of the equation of the hyperbola. If the transverse axis is parallel to the t/-axis the equation is raai (y - k) 2 (* - h) 2 _ ± [38J — - 2 p i- EXERCISES 1. Write the equations of the following hyperbolas: ' (1) Center at (4, — 3) , a =5,6= 3, and transverse axis parallel to oj-axis. (2) Center at (—6,-2), a = 2, b =4, and transverse axis parallel to y-axis. 2. Find the coordinates of the vertices and the foci, and the equations of the directrices of each hyperbola of exercise 1. 3. Find the equation of the hyperbola with center at (—2, 7), one directrix the line y = 5, and eccentricity equal to £ . Digitized by Google 144 ANALYTIC GEOMETRY [§115 4. Find the equations of the hyperbolas that are conjugate hyperbolas with those of exercise 1. 5. .Find the equations of the asymptotes of the hyperbolas of exercise 1 . 115. Equation of the form Ax 2 + Cy 2 + Dx + Ey + F - 0. Every equation of the form Ax 2 + Cy 2 + Dx + Ey + F = 0, where A and C have unlike signs, represents an hyperbola with axes parallel to the codrdinate axes. Proof. — In a manner identical to that of article 101 the equation takes the form ~T S1T\9 I A ET9 A A riTZ ™" *• CD 2 + AE 2 - 4ACF ' CD 2 + AE 2 - 4ACF 4tA 2 C 4AC 2 This is of the form of [38] or [38i] for the denominators have unlike signs since A and C are unlike in sign and therefore 4tA 2 C and 4tAC 2 are unlike in sign. If the second denominator is negative, the transverse axis is parallel to the X-axis. If the first denominator is negative, the transverse axis is parallel to the y-axis. From the preceding proof it follows that the equation of an hyperbola in the form Ax 2 + Cy 2 + Dx + Ey + F = can be transformed into the standard forms, [35] or [36], by a suitable translation of the coordinate axes, the new origin being at the point (-^Z' ~"2C/ Example.— Express the hyperbola 3&r* - 25y* + 21ftc + 100y-676 =0 in the form of [38]. What are the coordinates of its center, foci, and vertices; the lengths of the semi-axes; and the equations of its directrices and asymptotes? Plot. Finally, translate the codrdinate axes so as to change to the form [35] and answer the same questions with reference to the new axes. Solution. — Completing the squares in x and in y, 36(3* + Ox + 9) - 25(y* - 4y + 4) - 676 + 324 - 100, or 36(s + 3) 8 - 25(y - 2)* - 900. Dividing by 900 and putting in the form [38], . (s+3)* (y-2)» _\ 25 36 Digitized by Google Ins] THE HYPERBOLA AND CERTAIN FORMS 145 This is an hyperbola with its center at the point C(— 3, 2) and trans- verse axis parallel to the z-axis. The semi-axes are 5 an d 6. a 5 The distance from the center to the foci is ae =•- VST, and the foci are F(-3 +V6T 2) and *"(-3 - V$l, 2). The eccentricity e Fiq. 102. The vertices are 7(2, 2) and V'(-8, 2). The distance from the center to the directrices is VQ1 and the 5 5 equations of the directrices are x = — 3 H 7=. and x ■■ — 3 ;=• V81 V61 The asymptotes have slopes of $ and — # respectively, and pass through C(— 3, 2). Their equations are by [15], y - 2 - !(* +3). and y - 2 = -£(x + 3), or 6x - 5y + 28 - 0, and 6a; + by + 8 = 0. The hyperbola is as shown in Fig. 102. To change to the form of [36], put x + 3 = x 1 and y — 2 — yf. Then x'* y'* 1 becomes — — — = 1, referred 25 36 ' the equation (x + 3)» (y - 2)» 25 36 to CX< and. CY' as axes. The center is C(0, 0); foci are F(V6% 0) and F'(-\/6% 0); vertices are F(5, 0) and V(-5, 0); equations of directrices are x ■» ■ V6T 6* — 5y — and 6x + by 10 and a; — — \/61 ; and asymptotes are 0. Digitized by Google 146 ANALYTIC GEOMETRY [§116 EXERCISES 1. Express the equations of the following hyperbolas in the form of [88] or [88i]. Find the coordinates of the centers, foci, and vertices; the lengths of the semi-axes; and the equations of the directrices and asymptotes. Sketch each curve. (1) 9s* - 16y* - 108x + 96y + 36 - 0. (2) 16y* - x* - 6x - SOy + 75 - 0. (3) 8x* - 28y* - Sx - 2Sy - 61 - 0. (4) Sx* - 9y* - 16s + 64y - 1 - 0. (5) 3y* - 4s* - 16x - 24y - 52 *- 0. 2. Transform 9s* - 25y* + 54s + lOOy + 206 = by translating to new coordinate axes parallel respectively to the old axes, with new origin at (—3, 2), and sketch the curve. 8. Transform each of the hyperbolas of exercise 1 to the form of [86] or [36]. Find the coordinates of the foci, and the equations of the directrices referred to the new coordinate axes. 4. Find the equation of the hyperbola with conjugate axis parallel to the x-axis, center at the point (—3, 4), eccentricity f> and passing through the point (9, 4 + 8\/3). 6. Find the equation of the hyperbola whose axes are parallel to the coordinate axes and which passes through the points (3, 4), (—7, 4), (8, 4 + 4\/3), and (-12, 4 - 4\/3). 6. Find the equation of the hyperbola having a focus at (6, 2), a directrix the line s — 12 = 0, and e = 2. 116. Equation of hyperbola when axes are rotated. — In like manner to that for the parabola (Art. 90) and the ellipse (Art. 102), the equation Ax 2 + Cy 2 + Dx + Ey + F = 0, which is that of an hyperbola with axes parallel to the coordi- nate axes, is transformed by using equations [13] to the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. This is the most general form of the second degree equation in x and y r where B 2 -4AC>0. (See Art. 121.) Conversely f starting with an equation containing an xy-terin, rotation through a properly chosen angle will cause the xy-term to disappear by having its coefficient zero. EXERCISES 1. Transform the following equations by rotating the codrdinate axes through the angle given in each case: Digitized by Google §1171 THE HYPERBOLA AND CERTAIN FORMS 147 (l)s*-y» = 16. *--45°. (2) ~ - jj = 1. * - 90°. (3) sy - 8. *> = 45°. (4) 9y» - 16a;* - 144. <p - 60°. (5) x* - 4sy + y* + 9 = 0. <p - 45°. (6) a; 2 - 4sy + y* + 4\/2* - 2V% + 11-0. *> = tan"* 1. . 2. Transform the following equations into the standard form rotating the axes through a proper angle, and sketch the curve in each case: (1) x* + 4xy + y* - 16. (2) 9x* + U\/Sxy - 5y» - 48 - 0. 3. Simplify the following equation by first translating the axes to remove the x-term and the y-term, then by rotating through an angle that will remove the xy-term. Sketch the curve and the three sets of coordinate axes. x* + 2xy - y* + Sx + 4y - 8 - 0. Suggestion.— Find <p = 22J°. Then use sin 22J° = i y/2 - y/2 and cos 22i° - i V2 + V2. 4. In the hyperbola of exercise 3 find the coordinates of the center and the foci, and the equations of the transverse and conjugate axes, and asymptotes, referred to the original axes. 117. Equation of hyperbola in polar coordinates. — Here the procedure is similar to that for the parabola and ellipse, and the equations of these three conies should be compared. EXERCISES 1. Derive the equation in polar coordinates of the hyperbola with the pole at the left hand focus and polar axis along the transverse axis. Let p equal the distance from the focus to the directrix. 2 V Plot the following hyperbolas and draw the asymptotes: (1) p = 1 - e sin (2) p "" 1 + e sin 3. Transform x* + y* = e*(x + p} 2 into polar coordinates. 4. Show that in the equation of the hyperbola, p = ^ - -> ^ ' 1 — e cos the inclination of the asymptotes is cos"" 1 ( ±-\ • 6. Find the polar intercepts of the conic p = ^ -, and show * K 1 — e cos 6' that the transverse axis of the hyperbola is t y v and the major axis of the ellipse is 1 __ 2 » Digitized by VjOOQIC X48 ANALYTIC GEOMETRY [§118 6. Transform the polar equation p'cos20 — a 2 into rectangular coordi- nates, having the origin at the pole and the x-axis along the polar axis. 7. Show that p 2 -6* & 2 is an ellipse if e < 1; and that ■>*« 1 — e 2 cos 2 $ is an hyperbola if e > 1. Sketch each curve. 1 — e 2 cos 2 $ 118. Construction of an hyperbola. — First method. — The length of the transverse axis, 2a, and the foci F and F' are supposed known. On a drawing board place two tacks at F and F\ respectively, Fig. 103. Tie a pencil firmly at point P near the middle of a string. Pass one part of the string V l > ; i : x'A" o . a . / I Fio. 103. Fio. 104. under the tack at F and over the tack at F # , and the other part over the tack at F'. Adjust the string so that PF' - PF = 2a. Hold the parts of the string firmly together at Q and pull downward. The point P will generate an arc of an hyperbola. By arranging the string properly other arcs of the hyperbola may be generated. This construction depends upon the following. Theorem. — The difference of the distances from any point on an hyperbola to its two foci is constant and equal to the transverse axis. This may be proved as follows: In Fig. 104, from the definition of an hyperbola, PF' = e : N'P = e(* + f)= ex + a, and PF = e-NP = e(x - -) = ex - a. Digitized by Google §118] THE HYPERBOLA AND CERTAIN FORMS 149 Subtracting, PF' — PF — 2a = transverse axis. Second method. — A focus, the corresponding directrix, and the value of e are supposed known. In Fig. 105, let F be the focus, D'D the directrix, X'X the axis through the focus and intersecting D'D in A, and let the lines QR and TS be drawn through A with inclinations respec- tively equal to tan -1 ( + e). Also draw a series of lines parallel to D'D. Fig. 105. Then the points P and P' of the curve, on any one of these parallels, are found by striking arcs with the focus as center intersecting the parallel lines and using as a radius the length M N of that particular parallel. Show why this is so. EXERCISES 1. Locate a directrix and a focus and construct an hyperbola with e - $. With e - |. With e - \/3. 2. Construct a parabola by the same method. 3. Using the same method, construct an ellipse with (1) e «• }, (2) e - J V3, (3) e - f Digitized by Google 150 ANALYTIC GEOMETRY [§119 4. The difference of the distances of a point on an hyperbola from the foci is 4; and the foci are at the points (3, 0) and (—3, 0). Use the theorem of Art. 118 and derive the equation of the hyperbola. APPLICATIONS 119. Uses of the hyperbola. — Whenever the law connecting two variables is an inverse variation it gives rise to the equa- tion xy = fc, where x and y are the variables and k is a constant. This relation often occurs in physics, chemistry, and engi- neering. Boyle's Law which states that for a perfect gas the pressure varies inversely as the volume, gives rise to the equation pv = k. This is not used so much in practical work as is some slight variation of it. (See Art. 124.) Then again, if the law governing the location of a point is v such as to fulfill the conditions of the theorem of article 118 the locus is an hyperbola. These and other applications are best illustrated by examples. Example 1. — Given 20 c.c. of air at 1 atmosphere pressure. If the volume t; varies inversely as the pressure p, derive the equation showing the relation be- ptween the volume and the pressure. Plot the curve for values of p from 1 " ' atmosphere to 20 atmospheres. Fia. 106. Solution. — Since v varies inversely as the pressure, pv — fc. When p = 1, v = 20, hence 1-20 = fc, or k = 20. Therefore the equation showing the relation between p and t; is pv - 20. This is the equation of an equilateral hyperbola referred to its asymp- totes as axes, and can be plotted, as accurately as desired, by points. It is plotted in the first quadrant only because both volume and pressure must be positive. (See Fig. 106.) Digitized by Google §119] THE HYPERBOLA AND CERTAIN FORMS 151 V 1 2 4 6 8 10 15 20 V 20 10 5 3.3 2.5 2 1.3 1 Example 2. — Instruments for recording sound are placed at two points A and B 500 ft. apart, Fig. 107. The report of a cannon is recorded 0.25 second earlier at A than at £. Find the equation of the locus of the position of the cannon, and plot. (Sound travels 1120 ft. per second.) Solution. — Since sound travels 1120 ft. per second, the cannon is 0.25 X 1120 ft. = 280 ft. nearer A than B. Choose as origin the point midway between A and B, with x-axis through these points. Let P(x, y) be any position of the cannon. The coordinates of A and B are respectively (250, 0) and (-250, 0). Then BP - AP - 280. Fio. 107. OrV(-250 -aO* + y* - V(250 - *)» + y» - 280. Simplifying, = 1. 19,600 42,900 This is an hyperbola with a = 140, and b = 207.1. The conditions require, however, that the locus of the position of the cannon shall be the branch of the hyperbola nearer to A. Note. — The above example illustrates the principle made use of in the most accurate instruments used by the Allies in the Great War for locating hidden guns. Near the close of the war they were locating guns 10 miles distant within a radius of 5tf ft. EXERCISES 1. Given 10 c.c. of air at atmospheric pressure. If the volume v varies inversely as the pressure p derive the equation expressing the relation between the volume and the pressure. Plot for pressures from J atmosphere to 10 atmospheres. 2. Given an oak beam 10 ft. long of such dimensions that it supports 2 tons at its midpoint when resting at each end upon a support. If the weight w such a beam will support varies inversely as its length I derive the equation expressing the relation between w and L Plot for values of I from 1 ft. to 20 ft. Digitized by Google 152 ANALYTIC GEOMETRY [§119 3. Find the locus of the center of a circle tangent externally to two given circles. 4. Find the locus of the center of a circle haying one of two given circles tangent to it internally and the other tangent to it externally. 6. The base of a triangle is fixed, and the difference of the angles at the base is $*\ Find the locus of the vertex opposite the base. 6. Three instruments for recording sound are located at three points, A f B, and C, in a straight line. From A to B is 300 ft. and from B to C is 500 ft. A sound, such as the report of a cannon, is recorded at B 0.05 second after it is recorded at A, and at C 0.35 second after it is recorded at B. Find the location of the source of the sound in distance and direction from the point midway between A and B. Suggestion. — Choose the origin of coordinates at the point midway between A and B. Derive the equations of the hyperbolas and solve as simultaneous equations. Note that only one branch of each hyperbola is possible. The equations of the hyperbolas will be found to be 27.7s» - y* - 21716, and' {x - 400)* - 1.6y* = 38416. The solution of these equations gives x = - 70 and y - 337.7. GENERAL EXERCISES 1. Find the semi-axes, the eccentricity, and the coordinates of the foci of the hyperbola 2a? 1 — Zy* — 12. Also find the equation of the hyperbola that is conjugate with this. 2. Find the coordinates of the points of intersection of the hyperbola 2x* - Zy 1 = 12 and the circle x % + y 1 = 16. 3. Find the semi-axes, coordinates of foci, eccentricity, and equations of directrices of the hyperbola 9x* — 4y* — 54s + 16y + 209 « 0. 4. Show that the following equation represents two straight lines parallel respectively to the codrdinate axes: 12xy + 8z — 27y — 18 = 0. Transform the following equations as indicated, illustrating each by a drawing: 5. a* - lOxy +y*+x + y + l=0ti> 32s* - 48y* - 9. 6. x* - 2xy - y* - 2 - to x* - y* + V^ - 0. 7. Find the equation of the locus of a point that moves so that the difference of its distances from ( —4, 2) and (4, 2) is always equal to 8. x* y 2 8. Given the hyperbola — = 1, find the coordinates of the 25 16 point on the hyperbola, with abscissa double the ordinate. Digitized by Google §119] THE HYPERBOLA AND CERTAIN FORMS 153 9. Find the distances from the foci of the hyperbola -— — -— — 1 to a 25 16 point on the hyperbola, with abscissa 10. 10. Find the equation of an hyperbola whose axes are parallel respec- tively to the coordinate axes and which passes through the points (0, 0), (1,1), (-2, -1), and (-2, 2). 11. The lines x — 2y — and x + 2y — are the asymptotes of an hyperbola that passes through the point (—5, 3). Find its equation. 12. Prove that for all values of a the point (a sec a, b tan a) is on the x s y* hyperbola — — — = 1. a* b* 13. Prove that sec a is the eccentricity of an hyperbola with asymp- totes including an angle 2a. 14. Prove that the portion of an asymptote of an hyperbola, which is intercepted between the directrices is equal to the transverse axis. Digitized by Google CHAPTER IX OTHER LOCI AND EQUATIONS 120. General statement — In the previous chapters, for the most part, equations of the first and second degree in two variables, and their loci are considered. In the present chapter a consideration will be made of other equations also and their loci, where they are of importance in the study of more advanced mathematics, or are of use in immediate applications to science and engineering. Such equations and loci are of infinite variety and form. They may be divided into two classes, (1) algebraic' and (2) transcendental. Algebraic curves the degree of whose equations is higher than the second, and all transcendental curves that he wholly in a plane, are often called higher plane curves. In Cartesian coordinates an equation that can be expressed in a finite number of terms of the form Qx n y m , in which the variables are affected by constant exponents and Q is a con- stant, is called algebraic, all others are called transcendental. 121. Summary for second degree equations. — The most general equation of the second degree in two variables may be written in the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. Theorem. — In rectangular coordinates, the equation of the second degree in two variables represents a conic section. To prove this it is only necessary to show that, by a suitable change of the coordinate axes, the equation reduces to a form already discussed. Given Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. 154 Digitized by Google §121) OTHER LOCI AND EQUATIONS 155 From [13] substituting x = x' cos <p — y' sin <p f and y = x 1 sin ^> + y' cos ^>, A(a' cos <p— y'8in<p) 2 +B(x' cos<p—y' 8in<p) (x' sin ^>+y , cos^>) + C(z' sin ^> + y' cos ^) 2 + D(x' cos <p — y' sin ^) + 2?(z' sin ^> + y 1 cos ^) + F = 0. Expanding and collecting terms, x' 2 (A cos 2 <p + B sin <p cos ^ + C sin 2 ^>) + x'y'(—2A sin ^cos^ + 2C sin <p cos ^ — B sin 2 ^ + B cos 2 <p) + . j/ ,2 (A sin 2 <p — B Em <p cos <p + C cos 2 ^) + x'(D coa <p + E sin ^>) + y'{E cos ^ — D sin ^) + F = 0. In this equation the x'y 1 term will vanish if — 2A sin <p cos <p + 2C sin ^cos <p — B sin 2 ^ + B cos 2 ^ = 0. Or if B(cos 2 <p — sin 2 <p) = (A — C)2 sin ^ cos ^. By trigonometry, this becomes B cos 2<p = (A — C) sin 2^>. B [39] . • . tan 2* = j^zTq' Since the tangent of an angle may have any value from — a> to + °° > it is always possible to rotate the coordinate axes through such an angle that the zV-term will vanish. Further, since the smallest positive value of 2<p is less than 180°, <p is an acute angle. This value of <p can always be chosen for the rotation. The general equation then reduces to the form A'x 2 + C'y 2 + D'x + E'y + F' = 0. From the considerations of the previous chapters, this equation represents one of the conic sections as follows: (1) A circle if A' = C". (2) A parabola if A' ^ 0, E' ^ 0, and C" = 0, or if C" ?* 0, D' * 0, and A' = 0. (3) An ellipse if A' and C are of like signs and unequal. (4) An hyperbola if A' and C have unlike signs. Theorem. — The general equation o f the second degre e in x and y, Ax 2 + Bxy + Cj/ 2 _dr, Zta ,+ % $ X^ , mpni mtfaT Digitized by Google 4A- t i 156 ANALYTIC GEOMETRY (§12i p arabola? im -^Hfggg^or on hyperbola according a$ B 2 — 44. C fYpmfo awe, t'g fcafl lftan sero r or & flflflE dF %e*i g^rvT Proo/. — Using the values of A', £', and C", ii' = ii cos 2 ^ + Bsin^cos^ + C sin 2 <p, (1) 5' = B(cos 2 <p — sin 2 ^) — (A — C)2 sin ^ cos <p, (2) C — A sin 2 ^ — B sin ^ cos tp + C cos 2 <p. (3) Adding (1) and (3), A' + C" = A + C. (4) Subtracting (3) from (1), A' - C" = (ii - C) cos 2*> + £sin2*>. (5) Squaring (2) and (5) and adding, B' 2 + (A' - CO 2 = B* + U - C) 2 . (6) Squaring (4) and subtracting from (6), £' 2 - ±A'C = B 2 - 4AC. (7) But, if ^ is chosen so that tan 2<p = a _ n 1 &' ~ 0- Hence B 2 - 4AC = -4A'C. From this it follows that ' > (a) B 2 - 4AC - if either A' - or C = 0, \ (6) B 2 - 4AC<0 if A' and C" have like signs, (c) B 2 - 4AO0 if A' and C" have unlike signs. J r v " These are respectively the conditions necessary for a parabola, an ellipse, or an hyperbola. J M - *\ r v* Example,— -Given s 2 + 24a# - 0y* + 4s + 48y + 34 - 0. (1) Deter- mine whether it represents a parabola, an ellipse, or an hyperbola; (2) transform so as to free of the xy-term; (3) reduce to the standard form; (4) plot and show the three sets of axes. Solution.— (1) B» - 4AC = 24* - 4.1(-6) - 600. \ Therefore the equation represents an hyperbola. \ (2) tan 2<p - j^Tc " *+' cos 2<p " A * sin *> - yt(l - cos gg = V j(l - JV) - f. cos * - VJ(1 + cos 2*) - Vi(l + A) - f • ; Then the formulas [18] become x — |x' — f y 7 , and y — fa/ 4- f y'. Digitized by Google »22j: OTHER LOCI AND EQUATIONS 157 Substituting in the equation, {& - fyO* + 24(f* / - &)(%* + fyO - 6(4^ +t/)* +4(^ - f^) +48(Js' + !•) + 34 - 0. Simplifying, 10s" - 15Y» + 32^ + 36^ + 34 - 0. (3) Putting x 1 - x" + h and y* - y" -f fc, and simplifying, 10s"» - 15y"» + (20fc + 32)*" - (30fc - 36)y" + lOfc* - 15fc* + Z2h + 36fc + 34 - 0. Equating coefficients of x" and y" to 0, and solving, 20h + 32 - 0, 30fc - 36 - 0. .*. fc - -|, and k - f . Substituting these values and simplifying, 2*"» 3y"* + 6«0, or Y»^-= 1. This is an hyperbola with its center at the origin and its transverse axis along the y"-axis. (4) The three sets of coordi- nate axes and the curve are as shown in Fig. 108. Remark. — The values of h and k could have been found by completing the squares in x and y. In the solution given above, the rotation of axes was made first; but the work would have been shortened somewhat if the axes had been translated first. 122. Suggestions for simplifying second degree equa- tions. — If the equation is that of an ellipse or hyperbola, first translate the axes to remove the terms of the first degree, and then rotate the axes to remove the xy~term. If the equation is that of a parabola, first rotate the axes to remove the xy-term, and then translate to remove the constant term and one of the terms of first degree. It may-he. that the locus is a perrrtTGiat it" is composed of straight lines, or is imaginary. These forms are often called degenerate forms and are best discovered from the simplified E quation. — " ..- - Fio. 108. Digitized by Google 158 ANALYTIC GEOMETRY (§123 EXERCISES Test each of the following equations as to whether it is a parabola, an ellipse, or an hyperbola. Simplify each and plot showing all sets of , coordinate axes. > ,\., f[ A v t , ^ }/ / / 1. 6s* + 24sy - y* + 5% - 55 - 0. * ' 2. 25s* - Uxy + 25y* + 142s - 178j/ + 121 = 0. -\ 3. s* + sj/ + y* - 3y + 6 = 0.(2^ / 4. 32s* - 4Sxy + lSy* + 35s - 120y + 200 - 0. *V ; ' I 6. 13s* - 6\/5sy + 7y* - 64 - 0. , t 6. s 2 - 2y/Zxy + 3y* - 6\/3z - fy - 0. " ^ 7- 2s* + 6sy + 10y* - 2s - 6y + 19 - 0. 8. 6s* + 13sy + 6y* - 8s - 7y + 2 = 0» 9. 4s* + 4sy + y* + 4s - 3y + 4 - 0. 10. 9s* - 12sy + 4#* - 20s - 30y - 50 = 0. ALGEBRAIC EQUATIONS 123. Parabolic type. — Equations of the form y = oaf, where a is a constant and n is positive, are said to be of the parabolic type. I (1) When n = 2, y = ox 2 . The locus is the ordinary- parabola with its axis on the j/-axis, and has already been discussed. (2) When n = 3, y = ox 8 . The locus is called the cubical parabola. It has the form shown in Fig. 109, for a — 1. Discussion. — When s = 0, y = 0, and the curve passes through the origin. It is not symmetrical with respect to either coordinate axis, but is sym- metrical with respect to the origin. Why? For any positive value of s, y is positive; and for any negative value of s, y is negative. Hence the curve lies wholly in the first and third quadrants. This information together with a few points makes Fig. 109. it possible to sketch the curve with considerable accuracy. (3) When n = I, y = axi. The locus is called the semi- cubical parabola. It has the form shown in Fig. 110, for o - 1. *»x digitized by Google §124] OTHER LOCI AND EQUATIONS 159 Discussion. — When x = 0, y = 0, and the curve passes through the origin. Writing y = x* in the form y* — x z , it is seen that the curve is symmetrical with respect to the z-axis. For any positive value of x, y has two values numerically equal but opposite in sign. For any negative value of x, y is imaginary. Hence the curve lies wholly in the first and fourth quadrants. 124. Hyperbolic type. — Equations of the form y = ax n , where a is a constant and n is negative, are said to be of the hyperbolic type. (1) When n = — 1, y = aar 1 , or xy = a. The locus is the ordinary equilateral hyperbola lying in the first and third quadrants. (2) When n = —2, y = ax~ 2 y or x*y ^ a. The locus has the form shown in Fig. Ill, for a = 1. Fiq. no. Discussion.— No finite value of x will make y = 0, and no finite value of y will make x = 0. Hence the curve does not meet either of the coordinate axes. *-x =*•* Fia. 111. Fiq. 112. Since x is affected only by an even exponent and y only by an odd exponent, the curve is symmetric only to the j/-axis. For all positive finite values of y, x has two finite values equal numeri- cally but opposite in sign. For all negative values of y, x is imaginary. As y becomes large positively, x approaches zero both from the positive Digitized by Google 160 ANALYTIC GEOMETRY [§125 and the negative side. As x becomes large either positively or negatively, y approaches zero from the positive side. The curve lies in the first and second quadrants, and is asymptotic to both coordinate axes. (3) When n = — f, y = ax~l, or xty = a. The locus has the fdrm shown in Fig. 112, for a = 1. The discussion- is left as an exercise. EXERCISES Plot each group of the following equations upon the same set of coordinate axes, by first discussing the equation and then finding a few points. LY I 1. (1) V - x\ (2) y - x\ (3) y « x\ ~ ^ (2) !/-*«, (3)y-x*. (2) y=*«, (3)y«*«. (2)y=xl (2)y-ar». (2)y=ar«. (2)y = aT*. (2)y-ar«. 9. If p is the pressure and * the absolute *■ JC temperature of a gas in adiabatic expan- 7 sion, p = fct 7 "" 1 , where fc is a constant and y — 1.41 for air. If p « 2700 when t = 300, find fc, and plot the equation for values of t from 200 to 400. 10. In a mixture in a gas engine expand- ing without gain or loss of heat, it is found that the law of expansion is given by the equation pv 1 * 91 « c. Given that p = 188.2 when v =* 11, find the value of the constant c, and plot the curve of the equation using Consider values of v from 10 to 25. 1. (i) v = *», 2. (1) V = *, S. (1) V -»», 4. (i) y -*», 6. (1) V -*-', 6. Wv = *-«, 7. (1) V -=r*, 8. (i) y -*-«, Fig. 113. this value of the constant. 125. The cissoid of Diodes.— In Fig. 113, OT is the diameter of a fixed circle. At fa tangent is drawn, while about a secant revolves meeting the tangent in Q and the circle in R. The point P on the line OQ is taken so that OP = RQ. The locus of the point P is the cissoid of Diodes, Digitized by Google §126] OTHER LOCI AND EQUATIONS 161 To derive the equation of the cissoid, choose as origin and the x-axis along 07V- ^ Draw MP and NR perpendicular to OT. Denote the coordinates of P by (x, y). Let the radius of the circle be a. <- > \ From the definition of the cissoid, OP = RQ. And evidently OM = NT. But OM = x, hence NT - x and CW = 2a - x. Also NR is a mean proportional between ON and iVT. Hence NR = Vs(2a — x). ' v By similar triangles, OM : ON = MP : NR. Substituting values, x : 2a — x = y : Vs(2a — £)• x 3 . ' From this y 2 = ^ — ~ — > the equation required. The curve may be plotted from the definition given above, or from the equation. Note. — By means of the cissoid the problem of the dupli- cation of the cube can be solved. This problem, to find a cube that is double a given cube, was one of the famous problems of antiquity. 126. Other algebraic equations. — An unlimited number of definitions of loci could be given that would result in algebraic equations. There are many such curves that are of more or less historical importance as well as of value in mathematics and other sciences. Also there are an unlimited number of algebraic equations that may be discussed and their curves plotted. It should be remembered that an algebraic equation as truly defines a locus in terms of rectangular coordinates, as does the definition of the preceding article define the cissoid. 8a' Example 1. — Discuss and plot the equation y = x* +4a* Intercepts, — When x = 0, y — 2a. The curve does not meet the rr-axis, since no finite value of x will make y = 0. Symmetry. — Since only an even power of x occurs the curve is sym- metrical with respect to the y-axis. 11 Digitized by Google 162 ANALYTIC GEOMETRY [§126 Extent. — If a is a positive number, y is positive for all real values of x, and the largest value of y is 2a when x = 0. As x becomes very large in absolute value, y becomes very small but always positive. Hence the curve is asymptotic t o the s- axis in both directions. Or, solving for x, x » ±2a *l/ • Hence x is imaginary when <o. This is true when y<0 or when y>2a. Hence the curve lies in the first and second quadrants, is symmetrical with respect to the y-axis, and lies between the x-axis and the line y = 2a. Points on the curve and in the first quadrant can be found by choosing positive values for x. X a 2a 3a 4a 6a y 2a i« a *» i« & The curve is as shown in Fig. 114, and is known as the witch of Agnesi. Y « Fig. 114. EXERCISES Discuss the equations and plot the curves in exercises 1-16. 1. y - 2. y = 8 x -2 x + 3 9. xV - (V + 2) 8 (16 - y«). (Example of Conchoid of Nicho x -3 3. y = x(x — l)(x — 2). 4. y* = (x - l)(s - 4)(s - 6). x +3 (s - 2)(s + 1)* (a? - 4)(s + 3) (x + 1)<* - 2)' 7. x\ + y* = a*. 8. si 4- y* = <**• «. y 6. y 10. xy* = (* - a) 8 (2a - x). 11. 9y* = (s + 7)(3+4)«. 12. y = x 8 + x - 3. 13. y - *'+ 6s* + 10s - 2. 14. y = s< - lOz* - 4s + 8. 16 - * = FT* - 16. «*)■ + (Mi - I- Digitized by Google §127] OTHER LOCI AND EQUATIONS 163 • 17. Two fixed points F f and F are 2a units apart. Choose the origin at the center of the line joining F' and F, and the z-axis along this line. Find the equations of the loci of the point P(x, y) when FP (1) pfp — a constant not unity, (2) FP + F'P - a constant, (3) FP - F'P = a constant, (4) FP X F'P - a constant, k. In (4) the locus is called a Cassinian oval, and its equation is (** + y*)* - 2a 8 (s* - y*) - h* - a*. 18. Sketch the loci of (4) of the preceding exercise when a — 1, and k has successively the values 0, 1, and 2. 10. Write the equation of (4) of exercise 17 when k — a f , and plot the curve. This curve is called a lemniscate. 20. Express the lemniscate in polar coordinates, using the positive part of the z-axis as polar axis. 21. A uniform beam of length I, fixed in position by being held at one end, supports a weight at the other end. The deflection y at any distance x from the fixed end is given by the equation y — k(\lx* — Ja? 8 ). Find k for a beam 12 ft. long if the weight deflects the outer end 18 in., and plot a curve showing the shape of the beam for its entire length. Choose the fixed end as the origin and consider y positive when measured downward. TRANSCENDENTAL EQUATIONS 127. Exponential equations. — An equation of the form y = b x , where 6 is any positive constant, is called an expo- nential equation. If the exponent is fractional and involves even roots of 6, only the positive values of these roots are used. Example 1. — Discuss the equation y = b* when 6>1. Plot the curve when b = 1.5. Intercepts. — When x — 0, y = 6° » 1. This shows that the curve passes through the point (0, 1) for any value of b. If y = 0, b* = 0, which is impossible for any finite value of x. This shows that the curve neither meets nor crosses the rc-axis. However, for sufficiently large negative values of x, the value of b* can be made to become as near zero as desired. The curve is then asymptotic to the a>axis in the negative direction. Symmetry. — Since changing sto — 3 or y to — y changes the equation, the curve is not symmetrical with respect to either coordinate axis. Digitized by Google 164 ANALYTIC GEOMETRY [§128 Extent.— Since no integral value of a; can make y negative, and since only positive values of b* are to be taken when 2 is a fraction, the curve is wholly above the z-axis. Further, since y is not imaginary for any value of x, and increases as z in- creases, the curve lies in the first and second quadrants, exists for all values of x, and continually rises from left to right. x Plotting. — The curve of y m 1.5* can be plotted as accurately as desired by finding points. Taking logarithms of Fig. 115. both sides of the equation, log y = x log 1.5 = 0.1761a;. The following points are readily found, and the curve is as shown (1) (2) in(l) of Fig. 115. X -3 -2 -1 1 2 3 4 5 logy 1.4717 1.6478 T.8239 0.1761 0.3522 0.5283 0.7044 0.8805 y 0.296 0.444 0.667 1 1.5 2.25 3.375 5.063 7.595 Example 2. — Discuss the equation y = b* when 6<1. Plot the curve when b * J- The discussion is similar to that of example 1. It is to be noted that y decreases as x increases, and the curve is asymptotic to the z-axis in the positive direction. Plotting. — Points for plotting y = (J)* are found and the curve is as shown in (2) of Fig. 115. X -5 -4 -3 -2 -1 1 2 3 4 v 32 16 8 4 2 1 0.5 0.25 0.125 0.0625 128. Applications. — The most important case of the expo- nential equation is the case where the base is e, which is the base of the natural system of logarithms and equals 2.71828 • • < , It usually occurs in the form y = ae**, where Digitized by Google §129] OTHER LOCI AND EQUATIONS 165 a and k are constants, that may be determined in particular applications. This function is often called the "law of organic growth," or the "compound interest law," and is a function where the rate of increase or decrease at any instant is directly proportional to the value of the function at that instant. Just what the applications are cannot well be shown here, but the following uses are suggestive : (1) To express the pressure of the atmosphere at any height. (2) In physics and electricity, it is used in considering damped vibrations. (3) In medicine and surgery, to express the progress of the healing of a wound. (4) In biology, to determine the number of bacteria in a culture at any given time. (5) In chemistry, to express the progress of a chemical action. (6) In mechanics, in connection with the slipping of a belt on a pulley. Numerous applications will be discovered by the student as he progresses in his studies. Because of its frequent occurrence in problems involving conditions in nature, the base e is sometimes called ' ' a constant of nature." 129. Logarithmic equations. — The logarithmic equation is of the form y = logi, x, where 6 is a positive number different from 1. By the definition of the logarithm of a number, the equation y = log& x can be written in the exponential form x = b v . This is the same as the equation of article 127 with x and y interchanged. It is evident then that the discussion of the logarithmic equation y = log& x follows that of the exponential equation jc = b v } and gives the following when b > 1 : The x-intercept is at the point (1, 0). There is no ^/-intercept for as x approaches 0, y becomes — » , that is, the curve is asymptotic to the y-axis in the negative direction. Digitized by Google 166 ANALYTIC GEOMETRY [§129 The curve is not symmetrical with respect to either axis. When x > 1, y >0, and as x becomes » , y becomes «> also. When x < 1, y <0, and there is no value of y which will make x negative. Example 1. — Plot the curve of y = logio x. The following points are found, and the curve is (1) of Fig. 116. The unit on the y-axis is taken twice that on the x-axis. X 0.001 0.01 0.1 0.5 1 2 3 4 5 7 10 15 50 100 V -3 -2 -1 -0.301 0.301 0.477 0.602 0.699 0.845 1 1.176 1.699 2 Fig. 116. Example 2. — Plot the curve of y = log 2 x. The points can be readily found from the relation \ogiX = . s X log™ x = 3.322 log 10 x. lOgio A The curve is (2) of Fig. 116. X 0.001 0.01 0.1 0.5 1 2 3 4 5 7 10 15 100 V -9.97 -6.64 -3.32 -1 1 1.58 2 2.32 2.81 3.32 3.91 6.64 EXERCISES 1. Plot the following exponential equations: (1) y =- 2*. (2) y = 3*. (3) y = e*> where e - 2.718. (4) y = (0.75)*. (5) y - (0.4)*. 2. Discuss the effect upon the curve of y « 6* when 6< 1 and increases tol. Digitized by Google §130] OTHER LOCI AND EQUATIONS 167 8*. Discuss the effect upon the curve of y — b* when b> 1 and increases from 1. 4. Plot the curves of the following: (1) y = l-». (2) y - 2~». (3) y - *-, x>0. (4) y - «r*. 5. Plot the curves of the following: (1) y * log, x f where e - 2.718. (2) y - log* 2. (3) y - 2 log,*. 6. Discuss and plot the curve of y * — « — " Suggestion. — First plot yi » Je* and y% « Je~*. Then plot 6* -\-e~ 9 y » — 5 — by adding the ordinates j/i and yi to find y for the dfferent values of x. 7. Discuss and plot the curve of y » Ja(«o -f- «~ a ). This is the equation of the catenary, the curve assumed by a flexible cord sus- pended between two points. 8. A wire, weighing 0.2 lb. per foot, is suspended from two points in a horizontal line 50 ft. apart. The horizontal tension at each end is 10 lb. Plot the catenary formed by the wire. The constant a in the formula, y » Ja(e<* •+■ e~ a), is found by dividing the horizontal tension by the weight per unit length of the wire. 9. Plot the curve of i = ber**, where i and t are the variables. Choose b - 1.5 and a » 0.4. 10. If a body is heated to a temperature 7\ above the surrounding bodies, and suspended in air, its excess of temperature T above the surrounding bodies at any time, t seconds thereafter, is given by Newton's law of cooling expressed by the equation T — Tie""*, where a is a constant that can be determined by experiment. Given 7\ =■ 20 and a =- 0.014, plot a curve showing the temperature at any time t up to 100 seconds. 11. The dying away of the current on the sudden removal of the electro-motive force from a circuit containing resistance and self-induc- _ ?* tion, is -expressed by the equation, % = If L, where i is the current at any time, t seconds, after the e.m.f. is removed, R is the resistance, and L the coefficient of self-induction. Plot a curve to show the current at any time from t = to t = 0.2, if / » 10 amperes, #=0.1 ohm, and L = 0.01 henry. TRIGONOMETRIC EQUATIONS 130. The sine curve. — Discuss the equation y = sin x, and plot the curve. Digitized by Google 168 ANALYTIC GEOMETRY [§131 Intercepts. — When x = 0, y = 0. Hence the curve passes through the origin. When y = 0, sin x = 0, and x = mr radians, where n is any integer either positive or negative. Symmetry. — Putting — y for y or — x for x, changes the equation. Hence the curve is not symmetrical with respect to either axis. But putting — y for y and —x for x, does not change the equation. Hence the curve is symmetrical with respect to the origin. Extent. — Since there is a sine of any angle, the curve extends indefinitely in both the positive and negative directions. Since the sine of an angle is not greater than 1 nor less than — 1, the curve does not extend above the line y = 1 nor below the line y = — 1. Plotting. — Any length can be chosen as a unit on the coordi- nate axes. What may be called the proper sine curve is Fig. 117. plotted by choosing as a unit on the j/-axis the same length that is chosen to represent one radian on the x-axis. The curve is shown in Fig. 117. X .5 .707 ** i* br ** ** X br J* ** br *T br ¥* 2x y .866 1 .866 .707 .5 -.5 -.707 -.866 -1 -.866 -.707 -.5 From 2t radians to 47r radians or from — 2t radians to 0, these values repeat. They also repeat for each interval of 2w radians in both directions. 131. Periodic functions. — A curve that repeats in form as illustrated by the sine curve is called a periodic curve. The function that gives rise to a periodic curve is balled a periodic function. The least repeating part of a periodic curve is Digitized by Google §132] OTHER LOCI AND EQUATIONS 169 H* K* 2T +X Fig. 118. sin 2x. called a cycle of the curve. The change in the value of the variable necessary for a cycle is called the period of the func- tion. The greatest absolute value of the ordinates of a periodic function is called the amplitude of the function. In engineering and other practical applications of mathe- matics, there are many phe- nomena that repeat. It is for this reason that the periodic functions are of great import- ance. By a suitable choice of periodic functions almost any periodic phenomenon can be represented by a function. 132. Period and amplitude of a function. Example 1. — Find the period of sin nx, and plot y Since, in finding the value of sin nx, the angle x is multiplied by n before finding the sine, the period is — • . The curve for y = sin 2x is shown in Fig. 118. The period of the function is ir radians, and there are two cycles of the curve in 2x radians. Definition. — The number n in sin nx is called the period- icity factor. Example 2. — Find the amplitude of b sin x, and plot y — 2 sin x. Since, in finding the value of b sin x, sin x is found and then multiplied by b, the amplitude of the function is b, for the greatest value of sin x is 1. The curve for y — 2 sin x is shown in Fig. 119. The amplitude is 2. Definition. — The number 6 in 6 sin x is sometimes called the amplitude factor. By a proper choice of a periodicity factor and an amplitude factor a function of any amplitude and any period desired can be found. Fig. 119. Digitized by Google 170 ANALYTIC GEOMETRY [§133 133. Projection of a point having uniform circular motion. Simple harmonic motion. Example 1. — A point P, Fig. 120, moves around a vertical circle of radius 3 inches in a counter-clockwise direction. It starts with the point at A and moves with an angular velocity of 1 revolution in 10 seconds. Plot a curve showing the distance the projection of P on the vertical diameter is from at any time *, and find its equation. Plotting. — Let OP be any position of the radius drawn to the moving point. OP starts from the position OA and at the end of 1 second Pi tf . J P. Y P^ A Sf" rT 7 r -Jvj ?JL_^_^ H— f r^M i. o p Y>* ! 2 linn in t > ■ MOO 1 II ^ 1 i ' ' 1 i 1 k L__. —A 1... — ) * s 22 s'. 9 L__. fc-"i / p i \ Fig. 120. is in the position OP h having turned through an angle of 36° =» 0.6283 radians. At the end of 2 seconds it has turned to OP*, through an angle of 72° — 1.2566 radians, and so on to the positions OP it OPi t - • *, OP 10 . The points Ni, N%, • • • are the projections of Pi, Pj, • • • respect- ively, on the vertical diameter. Produce the horizontal diameter OA through A, and lay off the seconds on this to some scale, taking the origin at A, For each second plot a point whose ordinate is the corresponding distance of N from O. These points determine a curve of which any ordinate y is the distance from the center O of the projection of P upon the vertical diameter at the time t represented by the abscissa of the point. It is evident that for the second and each successive revolution, the curve repeats, that is, it is a periodic curve. Since the radius OP turns through 0.6283 radians per second, angle AOP - 0.6283* radians, and ON » OP- sin 0.6283*. Or y » 3 sin 0.6283*, the equation of the curve. Digitized by Google §133] OTHER LOCI AND EQUATIONS 171 In general, then, it is readily seen that if a straight line of length r starts in a horizontal position when time, t = 0, and revolves in a vertical plane around one end at a uniform angular velocity o> per unit of time, the projection y of the moving end upon a vertical straight line has a motion represented by the equation y = r sin <at. Similarly, the projection of the moving point upon the hori- zontal is given by the ordinates of the curve whose equation is y = r cos <at. The motion of the point N is a simple harmonic motion. If the time is counted from some other instant than that from which the above is counted, then the motion is represented by y =* r sin (<at + a), where a is the angle that OP makes with the line OA at the instant from which t is counted. As an illustration of this consider the following: Fig. 121. Example 2. — A crank OP, Fig. 121, of length 2 ft. starts from a posi- tion making an angle a = 40° = |r radians with the horizontal hne OA when t = 0. It rotates in the positive direction at the rate of 2 revolutions per second. Plot the curve showing the projection of P upon a vertical diameter, and write the equation. Plotting. — The axes are chosen as before, and points are found for each 0.05 second. The curve is as shown in Fig. 121. The equation is y — 2 sin (4rJ +#*•). Digitized by Google X72 ANALYTIC GEOMETRY [§134 Definitions.* — The number of cycles of a periodic curve in a Unit of time is called the frequency. It is evident that J — yF> where / is the frequency and T is the period. In y = r sin (at + a), f = jr- and T 7 = — ^71* to » The angle a is called the angle of lag. 134. Other applications of periodic functions. — The illustra- tions already given are by no means the only uses of periodic functions. Many uses occur in connection with sound, light, and electricity. Periodic curves are traced mechanically on smoked glass in experiments in sound and electricity. Such curves are also traced by instruments for recording heartbeats, breathing movements, and tides. Any periodic motion can be represented exactly, or can be closely approximated, by functions involving sines and cosines. 135. Exponential and periodic functions combined. — The" curve represented by the equation, y = be ax sin (nx + a), is important in the theory of alternating currents, in representing the oscillations of a stiff spring, the damped oscillations of a galvanometer needle, or the oscillations of a disk suspended in a liquid, such as is used to compare the viscosities of different liquids. The curve is most readily plotted by first plotting the curves represented by the exponential function and the periodic function separately on the same set of axes, and then finding the ordinates for various values of x by multiplying together the ordinates for these values of x in the exponential and periodic functions. It will be noted that the curve is periodic, and that the amplitude of the successive waves gets less and less while the wave length remains the same. Digitized by Google §136] OTHER LOCI AND EQUATIONS 173 Example. — Plot the curve showing the values of y for any value of x from x=— }*■ to a; = 2r f or the equation, y = e~°- 6 * sin (2x + Jx). The curve is readily plotted by first plotting j/i ■» e~°- oto and y s = sin (2a? + Jx), and then finding various values of y from the relation y = j/it/j. In Fig. 122, (1) is the exponential curve, (2) the sine curve, and (3) the final curve. Note that (3) and (2) intersect the x-axis at the same points. Fig. 122, EXERCISES 1. Plot y = sin x t using several different lengths on the z-axis as units. 2. Discuss and plot y = cos x. Give its period. 3. Discuss, and plot y = tan x, and y = cot x on the same set of axes. Give the period of each. 4. Plot y = sin x + cos x. Suggestion. — Plot f/i = sin x and y% — cos x on the same set of axes. Then find y from y = y\ + #a> by adding the ordinates for various values of x. 6. Plot y =* sin 1 x and y = cos 2 x on the same set of axes. 6. Plot y = sin"" 1 x and ?/ = cos" 1 re. 7. Plot y = sec a; and y = esc s, and give the period of each. 8. Plot y = sin \x } y = sins, y = sin2x, and y = sin fa; on the same set of axes. 9. Plot y = i sin s, y = sin x, y = 2 sin x t and y = f sin x on the same set of axes. Digitized by Google 174 ANALYTIC GEOMETRY [§136 10. Plot y = sin 2x + 2 cos x, and give the period. 11. Plot y — sin 3 + x. Is this periodic? 12. A crank 18 in. long starts from a horizontal position and rotates in the positive direction in a vertical plane at the rate of !*■ radians per second. The projection of the moving end of the crank upon a vertical line oscillates with a simple harmonic motion. Construct a curve that represents this motion, and write its equation. 13. A crank 8 in. long starts from a position making an angle of 55° with the horizontal, and rotates in a vertical plane in the positive direc- tion at the rate of one revolution in 3 seconds. Construct a curve showing the projection of the moving end of the crank in a vertical line. Write the equation of the curve and give the period and the frequency. 14. Plot the curves that represent the following motions: (1) y = 12 sin (1.88* + 0.44), (2) y - 2.5 sin (Jx* + T i*x). Give the period and frequency of each. 16. Plot y = rsin fat and y «■ r sin (Jx* + W" on the same set of axes. Notice that the highest points on each are separated by the constant angle Jx. Such curves are said to be out of phase. The difference in phase is stated in time or as an angle. In the latter case it is called the phase angle. 16. Plot y — r sin Jirf, y = r sin {\xt — Jx), and y = r cos Jxi all on the same set of axes. What is the difference in phase between these? 17. What is the difference in phase between the curves of y = sin x and y = cos x? Between y ■» cos x and y = sin (x + $*")? 18. Plot the curve y — e~* sin x for values of x from to 2x. 19. Plot the curve i « e~j* sin (2* -f- Jx) for values of t from —2 to 8. 20. In an oscillatory discharge of a condenser under certain con- ditions, the charge q at any time t is represented by the equation, q - 0.00224e-* 000 ' sin (8000* + tan" 1 2), where q is in coulombs and t in seconds. Plot the curve showing values of q for values of t from to 0.0012 second. What is the period? Suggestion. — Choose 0.0001 second as a unit on the t-axis, and 0.001 coulomb as a unit on the g-axis; and let the length representing a unit on the g-axis be about twice that for the unit on the t-axis. Plot the exponential curve first, and then the sine curve choosing as a unit on the g-axis the length representing 0.001 coulomb. EQUATIONS IN POLAR COORDINATES 136. Discussion of the equation. — As in the case of equa- tions in rectangular coordinates, in polar coordinates the dis- Digitized by Google §137] OTHER LOCI AND EQUATIONS 175 cussion of an equation helps greatly in learning the properties of the curve. The discussion is exactly similar to that in rectangular coordinates. (1) Intercepts. — (a) The intercepts on the polar axis are found by putting $ = 0°, 180°, 360°, • ••• nl80°. (b) The intercepts on the 90°-line are found by putting $ = 90°, 270°, etc. (c) Putting p =• and solving for 0, gives the values of $ for which the curve passes through the pole. (2) Symmetry. — (a) If the form of the equation does not change when — p is substituted for p, the curve is symmetrical with respect to the pole. (6) If it does not change when — $ is substituted for 0, the curve is symmetrical with respect to the polar axis, (c) If it does not change when ir — is substituted for 0, the curve is symmetrical with respect to the 90°-line. Show why each of these is true. Are their converses true? (See Art. 138.) (3) Extent. — If the equation is solved for p in terms of 0, the following can be determined: (a) Values of for which p has maximum or minimum values. In general this can be done readily when trigonometric functions are involved. (6) Values of for which p becomes infinite. These values determine the direction in which the curve extends to infinity, (c) Values of for which p is imaginary, that is, for which there is no curve. 137* Loci of polar equations. — Since some of the conditions of the previous article are sufficient but not necessary, care must be taken in determining symmetry and extent of curves. On the whole, however, the plotting is very similar to that in rectangular coordinates, and is best illustrated by examples. It will be found convenient to use polar coordinate paper. Example 1. — Discuss and plot p = 1 -f- 2 sin 6. Discussion. — (1) Intercepts on polar axis, = 0, p = 1; 9 ■* 180°, p - 1. Intercepts on 90°-line, 6 - 90°, p - 3; $ = 270°, p = -1. When p = 0, sin $ - -i, and $ - 210° or 330°. Digitized by Google 176 ANALYTIC GEOMETRY [§137 (2) Condition for symmetry with respect to the polar axis does not hold; but the curve is symmetrical with respect to the 90° -line since sin (x — 0) = sin 0. (3) Since p = 1 4- 2 sin 0, the maximum value of p will occur when sin = 1, or = 90°; and the minimum value of p will occur when sin $ = — 1, = 270°. No value of makes p imaginary. Plotting. — On account of the symmetry it is only necessary to find points for values of from 0° to 90° and from 270° to 360°. The curve is shown in Fig. 123. Fig. 123. sin P 0° 0.0 1.00 30° 0.50 2.00 45° 0,707 2.41 60° 0.87 2.73 90° 1.00 3.00 270° -1.00 -1.00 300° -0.87 -0.73 315° -0.707 -0.41 330° -0.50 0.0 345° -0.26 0.48 360° 0.0 1.00 Example 2. — Discuss and plot p = a cos 20. The four-leafed rose. Discussion. — (1) Intercepts on polar axis, — 0, p - a; B = 180°, p = a. Intercepts on 90°-line, = 90°, p = -a; = 270°, p = —a. When p - 0, B - 45°, 135°, 225°, 315°. (2) Symmetrical with respect to the polar axis, and the 90°-line. (3) Since p = a cos 20, the maximum values of p occur when cos 20 = 1, or when 0.= 0° and 180°. The minimum values occur when = 90° and 270°. Plotting. — On account of symmetry find points for values of in the first quadrant. The curve is as shown in Fig. 124. The arrow heads indicate direction in which the curve is traced as increases from 0° to 360°. Digitized by Google §138] OTHER LOCI AND EQUATIONS 177 e cos 20 P 0° 1.0 a 10° 0.94 0.94a 20° 0.77 0.77a 30° 0.50 0.50a 40° 0.17 0.17a 45° 0.0 0.0 60° -0.17 -0.17a 60° -0.50 -0.50a 70° -0.77 -0.77a 80° -0.94 -0.94a 90° -1.0 —a Fig. 124. 138. Remarks on loci of polar equations. — By definition, the locus of an equation requires that, (1) if the coordinates of any point satisfy the equation, the point is on the locus; (2) if any point is on the locus, the coordinates of this point satisfy the equation. In rectangular coordinates, there is no trouble in seeing that these conditions are fulfilled. This is because, in rectangular coordinates, there is one and only one point for every pair of coordinates; and, conversely, to every point there is just one pair of coordinates. In polar coordinates, trouble may arise since there is an ambiguity because a point has an indefinite number of pairs of coordinates determining it. Thus, in example 2 of the preceding article, it is seen that the point determined by the pairs of coordinates (\a } 60°), (-£a, 240°), (-Ja, -120°), and (Jo, -300°) is on the locus; but only (— Ja, 240°) and (— Ja, — 120°) satisfy the equation. For a like reason a curve may be symmetrical with respect to the polar axis even though (p, — 0) when substituted for (p, 0) changes the equation. In this case, some of the other pairs of coordinates of the point (p, — 0) would not change the equation. 12 Digitized by Google 178 ANALYTIC GEOMETRY [§139 *»x 139. Spirals. — Definition. — The locus of a point that revolves about a fixed point, and, at the same time, recedes from or approaches this point according to some law, is called a spiral. The fixed point is called the center of the spiral. When an angle is used in an equation and is not involved in a trigonometric function it is considered to be expressed in radians. Example.— Diacusa and plot the equation p = a # , p = a > 1. This is the logarithmic spiral. Discussion. — (1) When 0=0, p = l. (2) There is no symmetry. , p increases toward + » . As $ de- creases toward — », p approaches 0. Plotting. — The curve is readily plotted from a series of points. For a = 1.5 it is as shown in Fig. 125. Fig. 125. (3) As $ increases toward + < e -3 -2 -1 1 2 3 4 5 6 p 0.296 0.444 0.667 1 1.5 2.25 3.38 5.06 7.59 11.39 140. Polar equation of a locus. — The equation of a locus may often be f ound with greater ease in polar than in rectangular coordinates. The method is similar to that for finding the equation in rectangular coordinates, and has already been applied to the straight line and the conic sections. ° (See Arts. 69, 78, 91, 103, 117.) Example. — In Fig. 126, OT is the diameter of a fixed circle. At T a tangent is drawn, while about a secant revolves meeting the tangent in Q and the circle in R. The point P on the line OQ is taken so that OP = RQ. Find the equa- tion of the locus of P. Fig. 126. Digitized by Google §140] OTHER LOCI AND EQUATIONS 179 Solution. — Choose' as pole, OT as polar axis, and let OT « 2a. Let the polar coordinates of P be (p, 0). Then since ORT is a right angle, PQ « OR - 2a cos 0. Since OTQ is a right angle, OQ =* — ~ COS tt a>» a>^ »•* 2a rt A 2a sin* Hence p - OP = OQ — PQ — 2a cos 0, or p « — . y ^ ^ cos ' COS This may be transformed to rectangular ' codrdinates and obtain I/ 1 — o » the equation of the cissoid of Diodes derived in article 126. &Qj ~~% Compare the derivations of the equation by the two methods. EXERCISES ; Discuss and plot the following equations: *• p s f" Z a' ^ parabola. 2. p sin tan $ ^ 4a. A parabola. 3. p* cos 20 = a*. An equilateral hyperbola. 4. p = 3 cos + 2. Transform to rectangular codrdinates. 6. p = a tan* sec 0. Semi-cubical parabola. 6. p — a cot* esc 0, Semi-cubical parabola. 7. Transform equations of exercises 5 and 6 to rectangular codrdinates and compare with article 123. 8. p = a — 6 sin when a<b, when a = 6, and when a>b. Limacons of Pascal. 9. p* = a* cos 30. Is the curve symmetrical with respect to the 90°-line? Does the test apply? Sketch the following roses by first drawing the radial lines corre- sponding to values of which make p «= 0, and for values of which make p maximum in numerical value; and then determining the changes in the values of p between these successive values of 0. 10. p — a sin 20. 11. p — a sin 30. 12. p = a sin 40. 13. p = a cos 30. 14. p = a cos 40. 15. p — a cos 50. In plotting the curves of the following equations, it should be noted that in polar codrdinates it is sometimes necessary to carry the angle beyond 360° in order to secure the complete locus. 16. p — a sin 3 ^0. 17. p — a sin J0. 18. p* cos «= a* sin 30. 19. p = a(sin 20 -f cos 20). 20. p* — a* sin J0. 21. p = a(l ± cos 0). The cardioids. Discuss and plot the following spirals: . 22. p0 = a. Hyperbolic or reciprocal spiral. L ' 23. p = aO. Spiral of Archimedes. 24. p s =* a0. Parabolic spiral. Digitized by Google 180 ANALYTIC GEOMETRY [§141 25. />*$ =* a. The lituus or trumpet. * 26. Derive the equation of the locus of a point such that: (1) Its radius vector is inversely proportional to its vectorial angle. Ans. The hyperbolic spiral. (2) Its radius vector is directly proportional to its vectorial angle. Ans. The spiral of Archimedes. (3) The square of its radius vector is directly proportional to its vectorial angle. Ans. The parabolic spiral. (4) The square of its radius vector is inversely proportional to its vectorial angle. Ans. The lituus. (5) The logarithm of its radius vector is directly proportional to its vectorial angle. Ans. The logarithmic spiral. 27. Find the equation of the locus of the midpoints of the chords of the circle p = 2r cos 0, and passing through the pole. 28. Chords of the circle p = 2r cos $ and passing through the pole are extended a distance 2b. Find the equation of the locus of the extremities. PARAMETRIC EQUATIONS OF LOCI 141. Parametric equations. — When the coordinates of points on a locus are expressed separately as functions of a third variable, these equations are called the parametric equations of the locus. The new variable introduced in finding the parametric equations is called a parameter. The parameter may be introduced either for convenience or as a necessity, since in some cases it is easier to obtain the coordinates of points on a locus as functions of a third variable than it is to obtain a single equation connecting the coordi- nates of the points; and frequently two equations using the parameter can be obtained where it is not possible to obtain a single equation connecting the two variables. As will be seen, the parameter can be chosen in a great variety of ways, but it is usually chosen because of some simple geometric relation, or it is the time during which the point tracing the curve has been in motion. Example 1. — The parametric equations x = x\ + nt and y =» y x + mi represent the straight line which passes through the point (xi, y{) and has the slope — . n Digitized by Google \ ^ ; <-* X ^ - -f §141] OTHER LOCI AND EQUATIONS 181 That this is so can be seen by assigning values to t and plotting the values of x and y, or by eliminating t and obtaining the equation y y x - — (* - xi) f £(+*) *-x Fig. 127. which is the equation of a straight line. Example 2. — Consider a circle with center at the origin and radius r as generated by a point P starting on the x-axis and moving counter-clockwise. Then it is evident from Fig. 127 and the definitions of the sine and cosine, that the parametric equations x = r cos $ and y « r sin 0, ^ where is the angle generated by the radius to the point P, represent the circle. Also, squaring and adding the equations, X s + y* ■■ r*. Example 3. — The equations x = t* and y = 2t are parametric equa- tions of the parabola y* — 4a;, as can be seen by eliminating t from the two equations. The curve can be plotted by assigning values to t and computing the corresponding values of x and y. t - 4 -3 -2 -1 -i i 1 2 3 4 x 16 9 4 1 i i 1 4 9 16 y - 8 -6 -4 -2 -1 1 2 4 6 8 Fig. 128. The values of x and y are plotted, and the curve is as shown in Fig. 128. It is observed that as t varies from — » to + oo the corresponding point will trace out the curve, coming from » on the lower half and going to <» on the upper half of the parabola. Example 4. — The equations x=*a cos $ and y = b sin $ are parametric equa- tions of the ellipse as is shown in article 104. Example 5. — The equations x = a sec $ and y = b tan are parametric equa- tions of the hyperbola; for dividing Digitized by Google 182 ANALYTIC GEOMETRY [§142 the first by a, the second by 6, squaring, and subtracting the second from the first gives JC* y* ~i — fi = sec* — tan* 0*1. Example 6. — Equations (1) given in article 94, x = v cos at and y = v sin at — igt* t are parametric equations of a parabola. Here t is the number of seconds the point has been moving. EXERCISES 1. Write parametric equations of the straight line through (—3, 2) and having a slope of 2. f lot the line from these equations. 2. Write parametric equations of the circle with center at (2, 3) and radius 5. 3. Represent the parabola y* = 4a; by several pairs of parametric equations. Suggestion. — Either x or y can be represented at pleasure, but the other must be dete rmined in accordance with this. For instance, if x - t* + 1, y - 2 V** + 1. Plot the following parametric equations. In each case eliminate the parameter and find a single equation representing the same curve. 4. x = 4 - t\ y - t - 1. 5. x = 5 cos 0, y — 3 sin 0. 6. s — 2 4* sin 0, y = 2 cos 0. 7. s = e + %\ y = e - P. 8. 3 = 5+2 cos 0, y=4+3sin0. 9. x = 1 — cos 0, y = J sin J0. _10. s = cos 0, y = cos 20. 11. x = a sin -f & cos 0, y = a cos — b sin 0. , 12. x = a cos 1 0, y = 6 sin 8 0. 142. The cycloid. — The plane curve traced by a fixed point on a circle as the circle rolls along a fixed straight line is called a cycloid. The rolling circle is called the generator circle and the fixed straight line the base. The parametric equations of the cycloid can be derived as follows: In Fig. 129, let OX be the fixed straight line, C the generator circle of radius a, and P{x, y) the tracing point. Also suppose the circle is rolling towards the right. - Choose OX as the x-axis and the origin where the tracing Digitized by Google §143] OTHER LOCI AND EQUATIONS 183 point is in contact with the fixed line. Also choose as para- meter the angle 0, through which the radius to the tracing point turns. Draw the lines shown in the figure. Then x = ON - OM - NM = OM - PQ, and y = NP = MC- QC. [40] Fig. 129. But OM = arc MP = a$ f PQ = a sin 6, MC « a, and QC = a cos 0. Substituting these values gives x = a(8 — sin 8), y = a(l — cos 6). These are the forms of the equations most frequently used in dealing with the cycloid. If is eliminated the equation in x and y is x = a vers -1 - — y/2ay — y*, a form that is seldom used. EXERCISES 1. Plot the cycloid from the parametric equations. Is the curve periodic? 2. Construct a figure in which 9O°<0<18O°, and derive the equation of the cycloid from it. 3. Derive parametric equations for the locus traced by a point on a fixed radius and at a distance b from the center of the circle rolling as in generating the cycloid. First, suppose b<a; second, suppose b>a. 4. Plot the curves of exercise 3. Such curves are called trochoids. 143. The hypocycloid. — The plane curve traced by a fixed point on a circle as the circle rolls along a fixed circle internally is called an hypocycloid. Digitized by Google 184 ANALYTIC GEOMETRY [§U3 The derivation of the parametric equations is as follows: In Fig. 130, let be the fixed circle with radius a, and C the generator circle with radius 6. Let P(x, y) be the tracing point. Choose as origin and OX as x-axis. Also let the trac- ing point start at A where the x-axis intersects the fixed • circle. Choose as parameters the angle 6, through which the line of centers of the two circles turns, and the angle <p, through which the radius of the generator circle turns. Fig. 130. Draw the lines shown in the figure. Then x = OM - OB + BM = OB.+ NP, and y = MP = BN - BC - NC. But OB = OC cos 6 = (a - b) cos 6, and NP = CP sin PCN = b cos (<p - 6). Also BC = OC sin 6 - (a - 6) sin 0, and JVC = CP cos PCN = 6 sin (*> - 0). Substituting these values gives x = (a — b) cos + 6 cos (<p — 0), and y = (a — 6) sin — 6 sin ($> — 0). To eliminate the parameter <p } notice that arc AQ = arc PQ, or a0 = 6??, and hence ^ = ad Substituting the value of <p in the above equations, /a-b\ t«] x = (a — b) cos 8 + b cos ( — =- — j e, y = (a — b) sin 8 — b sin ( a 7" ) 6. Digitized by Google §144] OTHER LOCI AND EQUATIONS 185 Fio. 131. The hypocycloid is a closed curve only when the diameters of the two circles are commensurable. If o = 26, equations [41] become x = a cos 6 and y = 0. Therefore when the radius of the generator circle is one- half the radius of the fixed circle, the tracing point moves in a straight line. The most important special case of the hypocycloid is the four-cusped hypocycloid, in which a = 46. The curve is shown in Fig. 131. Here the parameter can be eliminated and a single equation in x and y obtained. Putting 6 = Jo in equations [41], x = fa cos0 + 1<* cos 30, and y = \a sin0 — \a sin 30. But from trigonometry cos 36 = 4 cos 8 6 — 3 cos 6, and sin 36 = 3 sin 6 — 4 sin 8 0. Substituting and simplifying, x = a cos 8 6 and y = a sin* 6. Affecting by the exponent | and adding, gives the equation in x and y, [42] x 1 + y 1 = a 1 . 144. The epicycloid. — The plane curve traced by a fixed point on a circle as the circle rolls along a fixed circle externally is called an epicycloid. Using a and 6 as the radii of the fixed circle and the generator circle respectively, and 6 and <p as shown in Fig. 132, the equations of the epicycloid are x = (a + b) cos e - b cos(?-^W [43] y - (a + b) sin 8 - b sin (— £^) *• An important special case of the epicycloid is the cardioid, in Digitized by Google 186 ANALYTIC GEOMETRY [§145 which a = b. The curve is as shown in Fig. 133. The equations here become x — 2a cos 6 — a cos 20, and y = 2a sin — a sin 20. P(*.V) r* Fig. 132. Fig. 133. P(*.v) +X 145. The involute of a circle. — If a string is wound around a circle, the curve in the plane of the circle^ traced by a point on the string as it is unwound and kept taut, is called the involute of the circle. The parametric equations may be derived as follows: Choose the x-axis through the point where the tracing point is in contact with the circle, and the origin at its center. The parameter 0, Fig. 134, is the angle through which the radius to the point of tangency of the string has turned. Then from the figure, x = OM = OB + BM = OB+LP, and y = MP = BL = BT - LT But OB = a cos 0, LP = TP sin 6 = aO sin $, and BT = a sin 0, LT = TP cos 8 = aO cos 6. Fig. 134. Digitized by Google §145] OTHER LOCI AND EQUATIONS 187 Substituting these values gives r^-, x = a cos 8 + ad sin 6 y = a sin 6 — ad cos 6. EXERCISES 1. Derive the equation of the f our-cusped hypocycloid. 2. Derive the equation of the epicycloid. 3. Derive the polar form of the equation of the cardioid from the parametric equations given in article 144. Suggestion. — In the polar form of the equation the pole is at A, Fig. 133. Notice that Z XAP .— Z XOC, and hence the parameter $ is equal to the polar coordinate 0. First, square and add the equations of Article 144, then translate to new origin at A (a, 0), finally, transform to polar coordinates and de- rive the equation p = 2a (1 — cos $). Digitized by Google CHAPTER X EMPIRICAL LOCI AND EQUATIONS 146. General statement. — In common every day affairs, in business, in the sciences as physics, chemistry, and biology, and in engineering, questions often arise involving the relations of variables. Values of these variables can be plotted accord- ing to some system of codrdinates, and, in this manner, curves obtained that give valuable information. Often the desired facts can be discovered directly from the curve; but frequently, especially in the sciences and in engineering, it is of the utmost importance to find a mathematical equation representing the curve more or less accurately. The determination of the equation may be a comparatively simple matter, but often it is very laborious and involves methods beyond the scope of this text. A curve that is plotted from observed values of the related variables is called an empirical curve or locus. The equation of an empirical curve is an empirical equation. Usually the empirical equation represents a curve that only approximates the empirical curve more or less accurately. 147. Empirical curves. — Innumerable examples of empiri- cal curves could be given. For many of these there may be no necessity nor reason for finding equations representing them. The rise and fall in the price of a certain stock may be represented graphically by using the price each day as the ordinate of a point of which the date is the abscissa. ( The curve drawn through these points win show at a glance the fluctuations of this particular stock. If the weight of a child is taken from month to month, a curve can be plotted by using the weights a3 ordinates and the corresponding dates as abscissas of points. 188 Digitized by Google §147] EMPIRICAL LOCI AND EQUATIONS 189 Empirical curves are often traced mechanically by instru- ments designed for that particular purpose. In this manner, at a weather bureau station, a curve is traced showing the relation between the temperature and the time. In Fig. 135, is a similar curve that shows the per cent of carbon dioxide Fig. 135. in the flue gas from a power plant. The variables are the time and the per cent of carbon dioxide. The system of coordinates is apparent. In Fig. 136, are plotted several curves showing the changes in the cost of living from July, 1914 to November, 1919. The data was taken from the Research Report issued by the National Industrial Conference Board. Digitized by Google 190 ANALYTIC GEOMETRY [§148 In such curves as these the information desired is gained directly from the curve, and no attempt would be made to derive an equation. 1914 1915 Year 1916 1917 1918 1919 Change in prices from 1014 to 1019. (1) Shelter; (2) Heat and light; (3) Sundries; (4) Cost of living; (5) Food; (6) Clothing. Fig. 13ft. 148. Experimental data. — In laboratory experiments and practical tests, pairs of simultaneous values of two varying quantities are measured. When these pairs of values are plotted, a curve is determined from which useful information may be obtained. The problem of finding the empirical equations representing such curves will now be considered. All data that is a result of measurements must be assumed to be subject to some degree of error, hence the endeavor will always be to approximate as closely as possible, both in the curve and in the equation. Sometimes in a problem of this kind the general form of the equation of the curve is known beforehand, and sometimes nothing at all is known but the coordinates measured in the Digitized by Google §149] EMPIRICAL LOCI AND EQUATIONS 191 experiment. If the general form of the equation is known, the computations for finding the definite equation can be made at once; but, if the general form of the equation is not known, the points are plotted so as to discover the general form if possible. 149. General forms of equations. — The f orms of equations frequently used are the following. Most of these have been studied in previous chapters, and should be reviewed, if necessary, so that their forms may be clearly in mind. (1) y = mx + by straight line. (2) y = cx n , n>0, parabolic type. (3) y = cx n , n<0, hyperbolic type. (4) y = ab x or y = ae* x , exponential type. (5) y = a + bx + ex 2 + dx* + • • • + qxP. For the parabolic type it is often necessary to use y — k = c(x — h) n , n > 0, where the vertex is at the point (A, h) ; and for the hyperbolic type y — k = c(x — h) n , n < 0. 150. Straight line, y = mx + b. — This is the form of the empirical equation when it is known that the relation between the variables is that of a direct variation. Since in the equation y = mx + b there are but two arbitrary constants, two pairs of measured values would be sufficient to deter- mine the equation completely provided the values could be measured accurately. Since this is not possible, a larger number of pairs of values are measured, and from these an equation is determined that represents the straight line lying most nearly to all the points. The method used in the follow- ing example for securing the equation ij called the method of least squares. The theory underlying the method is too difficult to be given here. Example. — Find the equation of the straight line that is in the form y — mx -f- 6, lying most nearly to the points determined by the follow- ing measured values of x and y: Digitized by Google 192 ANALYTIC GEOMETRY [§150 X 40 50 62.4 70 80.5 90 97 V 8.7 7.5 6.5 5.85 5.05 4.25 3.75 Solution. — Here the type of the equation is given so there is no need of plotting the points. Substituting each pair of values successively in the equation y — mx +b gives the seven equations: 8.7 = 40m +6, 7.5 - 50m + 6, 6.5 - 62.4m + 6, 5.85= 70m + 6, 5.05 - 80.5m + 6, 4.25= 90m + 6, 3.75= 97m + 6. Multiplying each of these by the coefficient of m in that equation and adding the seven resulting equations, gives 2690.875 = 36883.01m + 489.96. (1) r 15 10 10 20 80 40 60 60 Fig. 137. 70 80 90 100 Multiplying each of the seven equations by the coefficient of b in that equation and adding the results, gives 41.6 = 489.9m + 76. (2) Solving (1) and (2) for m and 6, gives m = -0.085, and b = 11.89. Substituting these values in y = mx «+■ 6, gives y = -0.085* + 11.89. (3) This is taken as the equation of the straight line lying most nearly to all the points. In Fig. 137 are plotted equation (3) and the points whose coordinates are the observed values of x and y. Digitized by Google §151] EMPIRICAL LOCI AND EQUATIONS 193 151. Method of least squares. — The method of least squares given in the previous article becomes tedious when there are many observations and the numbers are large. A sufficiently accurate result may be obtained by plotting the points, and obtaining the arbitrary constants of the equation by using two points that lie on the straight line that appears to be the best. If none of the plotted points lie on this line, use coordinates of points that do lie on the line. The method of least squares is quite mechanical, while the best straight line if determined by plotting is a matter of judgment and a good eye. The method by least squares for finding the empirical equation is stated in the following: Rule. — First, substitute each pair of observed values of the variables in the general equation. Second, if there are just as many equations as there are con- stants to be found, solve these equations for the constants. If there are more equations than there are constants, multiply each equation by the coefficient of the first constant in that equation, and add the resulting equations to form one equation. Proceed likewise for each other constant, and thus find as many equations as there are constants. Third, solve these equations for the constants. Fourth, substitute the constants, thus found, in the general equation and obtain the required empirical equation. EXERCISES 1. A wire under tension is found by experiment to stretch an amount I, in inches, under a tension T t in pounds, as given in the following table. Assume the relation I — kT (Hooke's law) and find the equation which best represents the relation between I and T. T 5 10 20 30 40 50 I 0.003 0.009 0.019 0.030 0.040 0.05S 13 Digitized by Google 194 ANALYTIC GEOMETRY [§151 2. Find the empirical equation in the form of y « mx + b best representing the relations between the values given in the following table: X 12 15.3 17.8 19 y 24.4 29 32.6 34.2 3. Find the equation of the straight line lying most nearly to the points determined by the following pairs of measured values: X 12 15 18 21 24 y 24.4 28.6 32.7 37.1 41.2 4. Find the -empirical equation connecting R and t from the following table of experimental values. R is in ohms and t in degrees centigrade. The equation is assumed to be in the form R = mt *+ b. t 10.1 15 21 26.8 33.1 40.4 R 9.907 9.923 9.940 9.959 9.979 10.002 5. Find the empirical equation giving H in terms of t, from the data of the following table. H is the total heat in a pound of saturated steam at t degrees centigrade. The general form is H = mt + b. t 65 85 100 110 120 H 626.3 632.4 637 640.9 643.1 6. In an experiment with a Weston differential pulley block, the effort Ej in pounds, required to lift a weight W, in pounds, was found to be as follows: w 10 20 30 40 50 70 90 100 E 3.25 4.875 6.25 7.5 9 12.25 15 16.5 Find the empirical equation in the form E = mW + 6. Digitized by Google §152] EMPIRICAL LOCI AND EQUATIONS 195 7. Plot the data given in exercise 6, and draw a line that, in your judgment, lies most nearly to all the points. Select two points that lie as nearly on the line as any, and determine the equation from the coordinates of these points. Compare the result with that of exercise 6. 8. In the following table, W is the weight of potassium bromide which will dissolve in 100 grams of water at t degrees centigrade. Find the empirical formula in the form W = mt + 6, connecting W and t t 20 40 60 80 w 53.4 64.6 74.6 84.'7 93.5 152. Parabolic type, y = ex 11 , n > 0. — If it is not known that the general form of the equation is of some particular type, it is well to plot the data on rectangular coordinate paper and judge the type from the curve. After the general form is selected, it is often difficult to determine whether or not it actually represents the observed values with sufficient accuracy for the purposes of the problem. A device that is of great assistance in determining whether to retain or reject the type selected is to transform the general equation into a linear equation, and see if the data plots as a straight line* This is done as follows when the equation is of the parabolic type: Given equation, y = cxP. Taking logarithms of both sides, log y = log c + n log x, which is a linear equation in log x and log y. If the points with coordinates (log x y log y), where x and y for each point are a pair of observed values, are plotted, and these points are found to lie approximately on a straight line, then the general form of the equation is suitable to the problem. The values of the constants, log c and n, can be determined by the method of least squares as in article 151. Of course, if the general form of the equation to be used is known to be of the parabolic type, the plotting is not necessary. Digitized by Google 196 ANALYTIC GEOMETRY [§152 Example.— Q is the quantity of water, in cubic feet per second, that flows through a right isosceles triangular notch when the surface of the still water is at a height H feet above the bottom of the notch. The values of H and Q in the following table are measured. Find the equa- tion connecting H and Q. H l 1.5 2 2.5 3 4 Q 2.63 7.25 15 26 41 84.4 Solution. — The values of H and Q are plotted in Fig. 138, and con- nected with the curve, which appears to be of the parabolic type. Assume the general form Q = cH n . (1) Taking logarithms, log Q = log c + n log H. (2) Put log H = x, log Q = y and log c = 6, and the equation becomes y » nx + b. Q Y a.6 .2 1.5 .1 Fig. 138. .2 .4 .6 .i Fig. 139. The values of x and y are found and plotted in Fig. 139. The points lie approximately in a straight line. logil 0.1761 0.3010 0.3979 0.4771 0.6021 log<? 0.4200 0.8603 1.1761 1.4150 1.6128 1.9263 Digitized by Google §163] EMPIRICAL LOCI AND EQUATIONS 197 Substituting the values for x and y in the equation y — nx + b, the following equations are obtained: 0.4200 - On +6, 0.8603 »0.1761n + fc, 1.1761 =0.3010n +6, 1.4150 = 0.3979n + 6, 1.6128 = 0.4771n + 6, 1.9263 =0.6021n +6. Solving these equations by the method of least squares, n - 2.5 and b - 0.4208. Substituting in equation (2), log Q = 0.4208 + 2.5 log H. Then log Q =* log 2.635 + log H*K Or log Q - log (2.635ff 1 *). .". Q — 2.635ff*- 6 , the required equation. The equation can be tested by computing values of Q for the several observed values of H f and comparing with the observed values of Q. 153. Hyperbolic type y = cx n , n < 0. — Data that are known to give an equation of this type can be handled in precisely the same manner as the parabolic type. The only difference that will arise Will be that the value of n is negative. 154. Exponential type, y = ab x or y =* ae 1 ". — The data from certain experiments, such as those involving friction, give rise to exponential equations. As with the other types the data can be plotted on rectangular coordinate paper and the general form of the equation determined. If it is thought to be of the exponential type, it can be tested by taking the logarithms of both sides of the equation and plotting on rectangular coordinate paper. If the points lie on a straight line, the assumed equation is correct. In order to express the form y = ab s in the form y = ae** it is only necessary to put 6 = 6*, whence log b = k log e, or k = j^Si = 2.3026 log b. loge Digitized by Google 198 ANALYTIC GEOMETRY [§154 Example. — From the following data determine the relation between W and 0. e 1.57 3.14 4.71 6.28 7.85 9.42 11 W 5.35 7.15 9.55 12.8 17.12 22.9 30.8 Solution. — First, plot the data given and determine the form of the equation to be used. The plotting is shown in Fig. 140, and the equation assumed is W - ab . (1) Second, to test this, take the logarithms of both sides of W = ab e . This gives log W = log a + B log b. Put log W = y, log a = B, and log b = m. This gives y = m$ + B. (2) W 30 20 10 —e O 5 10 Fig. 140. 1.5 / / 1 / .5 s* 9 O 5 Fig. 141. Arranging these values in a table and plotting, gives approximately a straight line as shown in Fig. 141. e 1.57 3.14 4.71 6.28 7.85 9.42 11 y 0.728 0.854 0.980 1.107 1.234 1.360 1.489 Digitized by Google §165] EMPIRICAL LOCI AND EQUATIONS 199 Substituting the pairs of values of $ and y in equation (2), 0.728 - 1.57m + B, 0.854 = 3.14m + B, 0.980 - 4.71m + B, 1.107 « 6.28m + B, 1.234 - 7.85m + B, 1.360 = 9.42m + B, 1.489= 11m + B. Solving these by the method of least squares, gives m = 0.0807 and B = 0.6005. Then a = 3.985 and b = 1.204. .'. W = 3.985 X 1.204 9 , the required equation. This expressed in the form W — ae** gives, W - 3.985e«. lw **. 155. Probability Curve, y = ae~ te *. — The curve that is perhaps the most widely used of any in dealing with experi- mental data is one variously called "the probability curve," "the error curve," and "the normal distribution curve." It is represented by the equation y = ae^ x \ where a and b are constants to be determined from the data. It is evidently symmetrical with respect to the y-axis. While definite uses of this curve are beyond the scope of this chapter, it may be stated that it is used wherever a most probable correct value is to be determined from a large number of independent measurements or observations. It is used in Digitized by Google 200 ANALYTIC GEOMETRY [§166 the study of statistics, in astronomy, biology, and chemistry, and in the study of theory of measurements. The form of the curve is shown in Fig. 142. 166. Logarithmic paper. — Because of the frequent occur- rence of formulas of the parabolic and hyperbolic types, con- siderable use is made in engineering practice of logarithmic paper, that is, paper that is ruled in lines whose distances, horizon- tally and vertically, are proportional to the loga- rithms of the numbers 1, 2, 3, etc. Logarithmic paper can be used instead of actually looking up the logarithms of the numbers as was done in the example of article 152. For if the values of H and Q are plotted as shown in Fig. 143, a straight line is de- termined just as when the logarithms of H and Q were plotted on rectangu- lar coordinate paper. Semi-logarithmic paper is ruled uniformly the same as ordinary coordi- nate paper in one direction, and in lines spaced as on logarithmic paper in the other direc- tion. Semi-logarithmic paper may be used to advantage when testing an exponential type. In Fig. 144, the values of 6 and W of the example of article 154 are plotted into a straight line. in 90 80 10 60 60 40 30 26 SO 16 10 9 8 7 6 6 4 8,5 3 ^~ ■ 2,5 2 1.5 1 1,5 2 2,5 3 4 5 6 7 8 9 Fig. 143. Digitized by Google §156] EMPIRICAL LOCI AND EQUATIONS 201 EXERCISES 1. Solve the equations of article 152 by the method of least squares and check the results given. 2. Determine the equation of the hyperbolic type connecting x and y from the following pairs of values: X 1.5 2.8 5.6 8.3 y 0.573 0.243 0.094 0.055 m w 18 70 60 60 40 80 ^ 20 10 9 ^9 8 r-* ** 7 6 5 n^ • 4 3 2 1 i 9 2 8 4 6 6 7 Fio. 144. 10 11 12 3. In propelling a ship of a certain class at 10 knots, the following pairs of values of D and H are measured, where D is the displacement in tons and H is the indicated horse-power. Find a formula of the parabolic type connecting D and H. Digitized by Google 202 ANALYTIC GEOMETRY t§156 D 1100 1530 1820 2500 3130 H 440 550 620 770 890 Compute H when D = 2000. 4. For different heights, h in feet above the surface of the earth, the reading, p in inches, of the barometer are taken as given in the following table. Determine a formula of the form p — cut* connecting p and h. h ' 886 2753 4763 6942 V 30 29 27 25 23 5. The data of the example of article 154 was taken in an experiment to determine the coefficient of friction /i, when a cord is wrapped around a cylindrical shaft, Fig. 145. In performing the experiment, the cord has a weight of 2 pounds attached to one end, and a pull of W pounds at the other end induces slipping when the arc of contact is radians. Determine the value of p for the equation W — aeP°. /i = & of Art. 164. 6. In testing the lubrication of certain oils in a bearing, 4} inches in diameter and 8 inches long with 250 revolutions per minute, the follow- ing pairs of values were measured, where p is the pressure in pounds per square inch and /i is the coefficient of friction. Determine a formula of the fprm n = ap n connecting p and /i* 1 p ; 65 115 215 315 465 'i 0.0090 ;0.0056 0,0036 .0.0028 0.0025 Plot the values showing that the curve is of the hyperbolic type. 7. In the same experiment as in exercise 6, but using another oil, the following values were obtained. Determine a formula connecting p and /*• V 65 115 215 315 415 515 M 0.00788 0.00528 0.00338 0.00267 0.00235 0.00215 Digitized by Google §157] EMPIRICAL LOCI AND EQUATIONS 203 8. In the following table are given the measurements taken in an experiment on friction, where /i is the coefficient of friction in a certain bearing running at a velocity of V feet per minute. Determine a formula of the form n — dV n connecting V and /i. V 105 157 209 262 314 366 419 471 M 0.0018 0.0021 0.0025 0.0028 0.003 0.0033 0.0036 0.004 9. In a mixture in a cylinder of a gas-engine, under adiabatic expan- sion, the following pairs of values are measured. Determine a formula in the form pv n = C connecting v and p. V 0.8 2 4 6 9 V 200 57 22 12.6 7.2 157. Empirical formulas of the type y = a + bx + ex 2 + dx 3 + • • • + qx n . — When a given set of corresponding pairs of values will not satisfy, in a satisfactory manner, any of the type equations already considered, the general equation y = a + bx + ex 2 + dx* + • • • + q& may be assumed. By substituting pairs of values in this equation, enough equations can be obtained to determine the constants a, 6, c, • • • . Since there must be, at least, as many equations as con- stants, no more terms can be assumed than the number of pairs of values measured. If there are more pairs of values than the number of terms assumed, the equations can be solved by the method of least squares. A less accurate method, but one more easily carried out, is to select as many of the equations as there are constants, and solve these for the constants. The equation thus found can be tested by substituting the pairs of values not used in the equations that are solved for the constants. If the points when plotted suggest a parabola, only three Digitized by Google 204 ANALYTIC GEOMETRY t§157 terms need be used. If the arrangement of the points is more irregular, more terms must be assumed. Example. — The following measurements at different depths were made to determine the rate of flow in a river, where x is the fractional part of the depth from the surface and y is the rate of flow. Determine a formula of the form y ** a + bx + ex 1 connecting x and y. X 0.2 0.3 0.4 0.6 0.8 0.9 y 3.195 3.253 3.261 3.252 3.181 3.059 2.976 Solution. — Substituting the pairs of values in y — a + bx + ex*, 3.195 -a + 06 + 0c, 3.253 = a +0.26 + 0.04c, 3.261 = a +0.36 + 0.09c, 3.252 -a +0.46 +0. 16c, 3.181 - a +0.66 + 0.36c, 3.059 - a +0.86 + 0.64c, 2.976 = a + 0.96 + 0.81c. These are solved by the method of least squares as follows: Multiplying each by its coefficient of a and adding the seven resulting equations, gives 22.177 = la + 3.26 + 2.1c. (1) Multiplying each by its coefficient of 6 and adding, gives 9.9639 - 3.2a + 2.16 + 1.556c. (2) Multiplying each by its coefficient of c and adding, gives 6.45741 = 2.1a + 1.5566 + 1.2306c. (3) Solving equations (1), (2), and (3) for a, 6, and c, a - 3.196, 6 - 0.438, c - -0.7608. Substituting these values in y = a + bx + ex 1 , gives y - 3.196 -«- 0.438* - 0.7608a; 1 , which is the required equation. EXERCISES 1. Plot the corresponding pairs of values of x and y given in the example of article 157, and draw a smooth curve lying as near as possible Digitized by Google §157] EMPIRICAL LOCI AND EQUATIONS 205 to all the points. Select the three points lying most nearly on the curve, and use the coordinates of these to find the values of a, b, and c. Com- pare with the result given in the solution by the method of least squares. 2. Determine an equation of the form y - a + bx + ex* + dx* from the following experimental values. Solve both by the method of least squares and by using the coordinates of four points. X 0.4 0.6 0.8 1.0 1.2 1.4 1.6 V 0.89 1.35 1.96 2.72 3.62 4.63 5.76 3. The melting point of an alloy of lead and tin containing x per cent of lead is t degrees centigrade. From the following table of measured values, find a formula in the form t = a + bx + ex*, giving the melting point of an alloy containing any known per cent of lead from 90 per cent to 35 per cent. X 87.5 84 77.8 63.7 46.7 36.9 t 292 283 270 235 197 181 For a further discussion of the subject of this chapter, the following works may be consulted: Merriman, Method of Least Squares; Weld, Theory of Errors and Least Squares; Johnson, Theory of Errors and Method of Least Squares; Palmer, Theory of Measurements; Steinmetz, Engineering Mathematics; Running, Empirical Formulas; Lipka, Cfraphr teal and Mechanical Computation. Digitized by Google CHAPTER XI POLES, POLARS, AND DIAMETERS 168. Harmonic ratio. — If two points A and B divide a line segment M N externally and internally in ratios that have the same numerical values, then A and B are said to divide MN harmonically. A and B are called harmonic conjugates with respect to the line segment MN. Theorem. — If the points A and B, divide the line segment MN harmonically , then the points M and N divide the line seg- ment AB harmonically. -e- M A N B Fig. 146. Proof— By hypothesis, j^-= - ^ MA AN Taking this proportion by alternation, ^=r= = ~"da?' Multiplying both sides of this equation by — 1, and replacing ~MA by AM and —BN by NB gives the required proportion AM _ AN MB NB 159. Poles and polars. — Definition. — If a line drawn through some point Pi is allowed to rotate about Pi while cutting a conic in the variable points M and N, then the locus of all points harmonically conjugate to Pi with respect to M and JVis called the polar of Pi with respect to the conic, and Pi is called the pole of the locus. 206 Digitized by Google §159] POLES, POLARS AND DIAMETERS 207 To find the equation of the polar of Pi with respect to an ellipse, suppose the ellipse in Fig. 147 is given in the standard form, a 2 + 6* If P 2 is the conjugate of Pi with respect to the ellipse and P1P2 is any line cutting the ellipse in the points M and N, then M and N are harmonic conjugates with respect to the line PiPj. If the coordinates of Pi are (xi, yi) and the coordinates of P 2 are (x 2 , y 2 ) ; then by [4] the coordinates of M are i y —z*f^ X \b pfs, **VT / ■h V N<r~ O V J*k Fig. 147. i r%X\ + rix 2 j r 2 yi + ny* \ \ ri + r 2 ' . ri + r 2 / and of N are ( r 2 3i — T&2 r\ — r 2 ny\ ri - r 2 / Since M and N are points on the ellipse, their coordinates must satisfy the equation of the ellipse, therefore / r 2 si + r x xt \ 2 V n + r 2 / + / r 2 yi + riy 2 \ 2 \ ri + r 2 / and / r 2 a?i - r x x% \ 2 / r 2 g/i - rig/ 2 \ \ ri — r 2 / , \ ri — r 2 / + = 1, - 1. o» 6 2 Clearing each equation of fractions and subtracting the second from the first gives the equation 4b 2 riT2XiX2 + 4a 2 rir 2 |/ij/ 2 = 4rir 2 a 2 6 2 . Dividing both sides of the equation by 4rir 2 and dropping Digitized by Google 208 ANALYTIC GEOMETRY (§159 the subscripts for the coordinates of the point ,Pj gives the equation of the polar of Pi b l X\X + a*yiy = a 2 6 2 . ' ' a* + 6 2 " x ' This shows that the polar of a point with respect to an ellipse is a straight line. If the point Pi is outside the ellipse the line drawn through Pi will not always intersect the ellipse. Algebraically the points of intersection of such a line and the ellipse have imaginary coordinates, but the coordinates of the point conjugate to Pi with respect to these points with imaginary coordinates are real. Hence that part of the locus obtained outside of the ellipse is also included as part of the locus. In like manner it can be shown that the polar of a point Pi with respect to the hyperbola a 2 b 2 l is XiX y x y _ t a 2 b 2 " l ' Also the polar of Pi with respect to the parabola y 2 = 2px is y\y = px + pxi. Likewise the polar of a point Pi with respect to the general conic Ax 2 + Bxy + Cy* + Dx + Ey + F = is [45] AxiX + |xiy+|xyi + Cyiy + 5 x + ^xi+?y + ^yi + F-0. The similarity should be noticed between this equation and the general equation of the conic written n in T\ 7} Jp w Axx +2 xy +"2 xy + C yy+2 X + 2 X+ 2 y + 2 y+Fts °' Digitized by Google §160] POLES, POLARS, AND DIAMETERS 209 These equations show that the polar of a point Pi with respect to any conic is a straight line. 160. Properties of poles and polars. — Theorem 1. — If two points are so situated thai one lies on the polar of the second, the second lies on the polar of the first. x 2 t/ 2 Suppose the conic is the ellipse -\ + r, = 1, and the point is Pi(xi, yj. Then the pokrofPiis^ + ^ - 1. If P2 lies on the polar of Pi its coordinates will satisfy that equation, hence a 2 - 1 " 6* l ' But this is precisely the condition that Pi shall satisfy the equation x& y*y _ , a 2 "*" 6 2 " l ' which is the equation of the polar of P 2 . This proof can easily be extended to the general equation of the second degree. Theorem 2. — If tangents can be drawn from a point to a conic, the polar of this point passes through the points of context of the tangents. In Fig. 147, the tangent Pi22 meets the conic in two co- incident points at R. Since the conjugate to Pi lies between these two coincident points, it must coincide with R. Likewise the polar must pass through S. Theorem 3. — Tangents to a conic at the points where a line cuts the conic pass through the pole of the line. This follows at once from theorem 2 in conjunction with the assumption that only one tangent line can be drawn to a conic at a given point. Example. — Find the pole of the line x + 2y = 1 with respect to the conic 3s* + 4y* — 6. 14 Digitized by Google 210 ANALYTIC GEOMETRY [§161 The polar of the point (x h y x ) with respect to the ellipse Zx* + 4y* = 6 is Zx x x + 4y\y = 6. Since this equation and x -\- 2y — 1 are equations of the same line the coefficients of x, y, and the constant term must be proportional, then TT and TT Hence X\ *= 2, y\ = 3, and the required pole is the point (2, 3). EXERCISES Find the equations of the polars of the points in exercises 1-8 with respect to the conies following. 1. (2, 3) 3s* + 4y* = 6. 2. (-1,6) 2s* + y* = -3. 3. (-1,2) 3s* -2y* = 1. 4. (1, -3) 2s* - 4y* - -5. 6. (1, 2) y* - 6s. 6. (-3, -2) s* = 4y. 7. (1, 2) s* - xy + 2/* - 6s - Zy + 2 - 0. 8. (3, -4) xy + Zy* + 3s + 1y + 1 = 0. Find the coordinates of the poles of the lines in exercises 9-16 with respect to the conies following. 9. 2s + 4y = 1, 6s* + 4y* - 3. 10. 2s + 2y - 1 - 0, 2s* + by 1 - 5. 11. 2s - Zy - 6 - 0, 4s* - Zy 2 = 12. 12. s - 2y + 4 = 0, y* - 2s - 0. 13. s + y + 1 = 0, s* + Qy = 0. 14. 4s + 5y = 2, s* + sy + y* - 3. 16. 3s + by + 2 - 0, s* + 2y* + s + y - 0. 16. 2s + 1 = 0, s* + 2sy + 2y - 2 - 0. 17. Find a point which with* (2, 4) divides the line joining (1, 1) to (4, 10) harmonically. 18. Prove that in any conic, the polar of the focus is the directrix. 161. Diameters of an ellipse. — Definition. — The locus of the middle points of a set of parallel chords of a conic is called a diameter of the conic. To find the diameter of an ellipse, let its equation be given in the form of [32] and suppose that the slope of the parallel chords is mi. Unless m\ is infinite, the equations of these chords have the form y = m\X + c, where mi is constant for Digitized by Google §161] POLES, POLARS, AND DIAMETERS 211 any one system of parallel chords, but c will have different values for different chords of the system. Suppose MN, Fig. 148, is one of these chords, the coordi- nates of M and N can be found by f solving simultaneously the equations y = m\X + c x t y 2 ' t\\\>Vft\Yl ** and Si + gi- h Eliminating y between these equa- tions gives Fig. 148. (6 2 + a*mi 2 )x* + 2a 2 cm x x + a 2 c 2 - a 2 b 2 = 0. The two roots of this equation are the abscissas of the points M and N. Half their sum is the abscissa of Pi(xt,y*), the middle point of MN. By a well-known theorem, Art. 4, the sum of the roots of the quadratic equation Ax 2 + Bx + C = S is equal to — -r- Hence a 2 cmi 6 2 + a 2 m! 2 ' To find t/2, substitute this value of x% in the equation y = mix + c. Then y. = 6 Tqrjfi^- The relation between x% and y% for any one, and therefore for every one, of these parallel chords must be independent of c. Hence eliminate c by dividing y% by x^. This gives Hi = ft 2 Xi a 2 mi Dropping the subscripts for P 2 gives the following equation of the diameter which bisects all chords of slope mil b 2 y = — 5 — x* Digitized by VjOOQlC 212 ANALYTIC GEOMETRY [§162 This is the equation of a straight line passing through the center of the ellipse. If mi is infinite, the parallel chords are all parallel to the y-axis and the symmetry of the ellipse shows the x-axis to be the diameter. If mi = 0, the parallel chords are all parallel to the z-axis, and the symmetry of the ellipse shows the y-axis to be the diameter. Since mi can have any value, any line passing through the center of the ellipse is a diameter. The length of a diameter of an ellipse is the distance be- tween the points where the diameter cuts the ellipse. 162. Conjugate diameters of an ellipse. — The slope of the diameter bisecting all chords parallel to the diameter is -6* \a 2 m\I and its equation is y = mix (2) But the diameter (2) is the diameter of the ellipse parallel to the set of parallel chords of article 161. Hence the diameter (1) bisects all chords parallel to diameter (2), and the diam- eter (2) bisects all chords parallel to diameter (1). Two diameters such that each bisects all chords parallel to the other are called conjugate diameters. Hence diam- eters (1) and (2) are conjugate diameters. If m 2 is the slope of (1) 6 2 or 6 2 mim2 = — --=• a' Digitized by Google §163] POLES, POLARS, AND DIAMETERS 213 163. Diameters and conjugate diameters of an hyper- bola. — Methods exactly similar to those in articles 161 and 162 show that all diameters of an hyperbola pass through X 2 V 2 its center. The diameter of — 9 — j- 9 = 1, which bisects all a 2 b 2 ' chords of slope mi, Fig. 149, is V = b 2 +*x Fia. 149. The slopes of two conjugate diameters of an hyperbola are connected by the relations b 2 ar The length of a diameter of an hyperbola when the diameter meets the hyperbola is the distance between the points where the diameter cuts the hyperbola. If the di- ameter does not cut the hyperbola, its length is defined as the distance between the points where it cuts the conjugate hyperbola. 164. Diameters and conjugate diameters of a parabola. — Let the slope of the parallel chords be mi, Fig. 150, and let their equations be y = m\X + c, where c will have different values for different chords. The ordinates of the points of in- tersection of the parabola y 2 = 2px, and these parallel chords are given by the equation m l2 / 2 - 2py + 2pc = 0. If y% is the ordinate of any one of their middle points V V 2 = ™ mi Since this equation is independent of c, it is the condition Fia. 150. Digitized by Google 214 ANALYTIC GEOMETRY [§165 that all points on the diameter must satisfy. Hence dropping subscripts, the equation of the diameter of a parabola is V This shows that the diameter of a parabola is a straight line parallel to its axis. Since mi may have any value except 0, any line parallel to the axis of a parabola is a diameter. As mi approaches 0, the system of parallel chords approaches parallelism to the axis of the parabola, and y increases without limit. Hence the diameter bisecting chords parallel to the axis of a parabola does not he in the finite part of the plane. 165. Diameters and conjugate diameters of the general conic. — Since the slope of a line remains unchanged by trans- lation of axes, all the results obtained so far hold good after translation for conies whose axes are parallel to the coordinate axes, providing that in the ellipse and hyperbola the major and the transverse axis respectively, and in the parabola the axis of the parabola are parallel to the x-axis. Formulas obtained for conjugate diameters, and equations of diameters do not hold true for rotation of axes unless account is taken in mi and m 2 of the change made by the rotation. EXERCISES 1. Find the equation of the diameter of the ellipse 3s* + 4y* = 6, which bisects chords of slope 3. Chords of slope — J. 2. Find the equation of the diameter of the hyperbola 2x* — 4y* = l t which bisects chords of slope 3. Chords of slope — J. 3. Find the equation of the diameter of the parabola y* — 4z, which bisects chords of slope 3. Chords of slope — J. 4. Find the equation of the diameter which bisects chords of slope 3, for the ellipse 2x* + Zy* - 4x - 12y + 2 = 0. Suggestion. — Translate axes to center of conic, and then translate back to the original axes. 6. Find the equation of the diameter which bisects chords of slope 3, for the hyperbola 2s* - 3y* - 4a; + 12y - 22 - 0. Digitized by Google §165] POLES, POLARS, AND DIAMETERS 215 6. If (2, 1) is one extremity of a diameter of the ellipse 4a?* + 9y* = 25, find the coordinates of the extremities of the conjugate diameter. 7. If the point (1, 2) is one extremity of a diameter of the hyper- bola 25s* — 4y* = 9, find the coordinates of the extremities of the con- jugate diameter. x s y* 8. If (x\, yi) is an extremity of a diameter of the ellipse ~i + m = 1, what are the coordinates of the extremities of the conjugate diameter? 9. Prove that the sum of the squares of any two semi-conjugate diameters of an ellipse is constant and equal to a 1 + b*. x 2 v* 10. If (xi, yi) is an extremity of a diameter of the hyperbola -^ — jk = 1, what are the coordinates of the extremities of the conjugate diameter? 11. Prove that the difference of the squares of any two semi-conjugate diameters of an hyperbola is constant and equal to a* — 6*. 12. Find the equation of the chord of the hyperbola 2a?* — Zy* = 6, through the point (4, 1) which is bisected by the diameter y = 4x. 13. Find the equation of the chord of the ellipse x* + 2y* = 4, through the point (6, 3) which is bisected by the diameter Zy + x - 0. 14. Find the equation of the chord of the parabola y % = 4x through the point (1, 6) which is bisected by the diameter y — 3. 15. Prove that the polar of any point Pi(a?i, y\) on a diameter of an ellipse is parallel to the conjugate diameter. 16. Two lines connecting a point on an ellipse with the ends of a diameter are called supplemental chords. Prove that supplemental chords are always parallel to a pair of conjugate diameters. 17. Prove that if a parallelogram is inscribed in an ellipse its sides are parallel to conjugate diameters. 18. Find the locus of the middle points of chords which connect the ends of pairs of conjugate diameters of a fixed ellipse. Digitized by Google CHAPTER XII ELEMENTS OF CALCULUS 166, Introductory remarks. — As has been stated, the discovery of the methods of analytic geometry during the first half of the seventeenth century gave the first great start in the development of modern mathematics. During the latter half of the same century Newton and Leibniz, building upon the writing and teaching of Isaac Barrow and others, discovered the method of the infinitesimal calculus. In this subject are studied very powerful methods of investigating functions and problems concerning variables. It is in the calculus that we find the greatest development of mathe- matical analysis and its applications in almost every field of science and engineering. Some of these methods and applications will now be considered. Here, as is always the case in the study of mathematics, it is necessary to understand clearly what is under consideration and how it is represented in mathematical symbols. 167. Functions, variables, increments. — Example 1. — If a suspended coiled wire spring has a weight attached to its lower end, the spring will be stretched. The amount of stretching will depend upon the weight, the greater the weight the greater the elongation. The elongation is then a function of the weight. If the weight is not so great that the elastic, limit of the spring is exceeded, the elongation varies directly as the weight The law connecting the variables is then stated by the linear equation V = kz, where y is the elongation, x the weight, and k a constant. 216 , Digitized by Google §167] ELEMENTS OF CALCULUS 217 That is, y is a function of x, and a change in the variable x produces a corresponding change in y. A change in the weight is called an increment of the weight, or an increment of x, and is represented by the symbol Ax (read "increment of x yy or "delta x"). A corresponding change in the elongation is called an increment of the elongation, or an increment of y, and is represented by Ay. Here x represents the independent variable and y the de- pendent variable. It is evident that for every Ax there is a Ay. Their relation may be shown as follows: (1) (2) Ax N Fig. 151. For any particular value of x as x x , y x = kx x . If x = Xi + Ax, y x + Ay = k(x x + Ax). Subtracting (1) from (2), Ay = kAx. That is, Ay varies directly as Ax, and is independent of the value of x. This is shown graphically in Fig. 151. The locus of y = kx is a straight line with slope k. Pi is a point on the line with coordinates (xi, yi). MN = PiQ = Ax, and QR - Ay. No matter what the magnitude of Ax, Ay = Ax tan QPiR = kAx. Example 2. — The distance s that a heavy body near the earth's surface falls from rfest in time t is given by the formula * 8 = hgt 2 . If t - t l9 Sl = \gt]. , (1) If t = h + A*, 8i + As = Mh+ A*) 2 . (2) Subtracting (1) from (2), As = ^(2*^+ At 2 ). That is, the value of As depends upon both t and At. This is shown graphically in Fig. 152. The locus of s = igt 2 is a parabola. The point Pi has coordinates (h, si), and Digitized by Google 218 ANALYTIC GEOMETRY [§167 At and As are as shown in the figure. It is evident from the figure that As depends upon both t and At. Definitions and notation. — If y is a function of x it may be written y = f(z), which is to be read "y equals a function 'y equals / of a." For convenience other symbols may be used for functions, as F(x), <p(x), /'(*)> etc. In the equation y = f(x), that is, when the equation expresses y explicitly in terms of x, y is an explicit function of x. If two variables are involved in an j+t equation in such a manner that it is necessary to solve the equation in order to express either explicitly in terms of the other, then either variable is said to be an implicit function of the other. Thus, in x 2 + y* = r*, y is an implicit function of x and x is an implicit function of y. If this is solved for y, y = ±\/r 2 — x 2 , in which y is an explicit func- tion of x. If solved for x, x «■ ±\Zr* — y 1 , in which x is an explicit function of y. Implicit functions of x and y may be written /(x, y), F{x, y), ip(x y y), etc. In the same discussion or problem the same functional symbol is used to represent the same function. Thus, if f(x) - 2x* + Zx + 1, then /(a) = 2a* + 3a + 1, and /(3) = 23* + 33 + 1= 28. If /(*> y) = 3s* + 4xy - y, then /(2, 3) - 3-2* + 4-2-3 - 3 = 33, and f(y, x) = Zy* + \xy — x. EXERCISES 1. If y = 10g and x x is any particular value of x, find Ay when x takes the increment As. Find Ay when Xi = 4 and Ax = 2. Find Ay for any other value of as and Az = 2. Plot so as to show these graphically, Digitized by Google §167] ELEMENTS OF CALCULUS 219 2. If y — 2x* + 1 and x x «■ 2, find Ay when As = 0.5. Find Ay when Ax - 0.01. Plot. 3. Express the area A of a square as a function of its side x. Find AA for x = 6 and Ax = 1. Illustrate by means of a square. 4. Express the area A of a circle as a function of its radius x. Find AA f or x — 10 and Ax =0.5. Illustrate by means of a circle. 6. Express the area of a square as a function of its diagonal. Express its diagonal as a function of its area. 6. Express the circumference of a circle as a function of its area. Express the surface of a sphere as a function of its volume. 7. Express the volume of a right circular cylinder as a function of its radius and altitude. Express the altitude as a function of its volume and radius. Express its lateral area as a function of its volume and diameter. 8. If /(*) - x* + 3s»- 2x - 4, find /(0), /»),/( -4). 9. If F{x) - y/x^+T, find F(0), F(-3), F(W$). 10. If <p{x) - logic x, find *>(100), ?(47.62) f ^(0.012). 11. If /(*) - cos *, find /(30°), /(**•), /(240°). 12. If/(s,y) - Zx*y + 4sy* - 2y\ find/(-s,y),/(ar, -y)./(-s, -y). 13. If /(y) - 3*, prove /(s)-/(y) -/(* + y). 14. If/(x) = sinsandFfc) = cos a;, prove that /(*+y) =/(^(y)+F(x)/(y). 15. If y = sin a;, express x explicitly in terms of y. If y = 2*, express x explicitly in terms of y. In each of the following equations express each variable explicitly in terms of the other, if it can be done by methods previously studied. 16. ~ + £= 1. 23. . ToTXTl - e ~*'- a* b 2 sin (2t + i*-) 17. a* + y* - a*. 24. *>* cos 20 - a*. 18. as* + y* = a*. 25. ^ sin tan $ = 4a. 19. 3(3 — 2a)* — ay* = 0. 26. *>* cos $ = a 2 sin 30. 20. «y + 4s 4 = 16. 27. logios— logi y+ 3 log™ a =0. 21. 4x* + y* - 8x - 2y + 1 = 0. 28. sin" 1 x - sin" 1 y = 45°. 22 *V _ /„ + 2 )t 29 * V - *~ 1 22 * 16 - y* - {y + 2) * 39 ' *« - 1 - * + y 30. If 8 - 16**and*i - 2, find As and -^ when A* - 1; when A* - 0.1; As when At = 0.01: when At - 0.001. What value does -tt seem to be ' At approaching as At becomes smaller? At/ 3L If y = x % and X\ = 1, find Ay and -^ when Ax = 10; when As = 1; Digitized by Google 220 ANALYTIC GEOMETRY [J168 when As — 0.1; when Ax — 0.01; when Ax — 0.001. What value does -~ approach? What then is the slope of the tangent at the point where x — 1 of the curve of y = x»? Plot. LIMITS 168. Illustrations and definitions. — Considerable use has been made of limits in elementary geometry, trigonometry, and algebra, but much greater use is necessary in the study of calculus. The following are simple examples of limits: (1) The variable which takes the successive values 1.3, 1.33, 1.333, • • • has as a limit \\. That is, the more figures there are taken, the more nearly the number approaches 1$. (2) The number \/2 is the limit of the successive values 1.4, 1.41, 1.414, 1.4142, • • . The diagonal of a unit square is the limit of the line lengths represented by this series of numbers. (3) If a point starts at the end A of the line AB, Fig. 153, and during the first second moves half the length of the line to C; during the next second, half — | — |»m of the remaining distance to D; continuing in this way to move Fia. 153. i ii. ,i . . ,. , i half the remaining distance dur- ing each successive second, then the distance that the point is from A is a variable of which AB is the limit. 12 (4) If y = — -t-^ and z is a variable approaching 2 as a limit, then evidently y is a variable approaching 3 as a limit. Definitions. — When a variable changes in such a manner that its successive values approach a constant so nearly that the difference between the constant and the variable becomes and remains less, in absolute value, than any assigned posi- tive number, however small, the constant is the limit of the variable. The variable is also said to approach the constant as a limit. If the variable is represented by x and the constant by a, then 4- Digitized by Google §169] ELEMENTS OF CALCULUS 221 the statement "x approaches a as a limit " is written thus, x = a. lim The form ^^^ [f(x)] = A is read "the limit of f(x) as x X "~ a approaches a as a limit is A." When a variable changes in such a manner that it becomes and remains greater than any assigned positive number, however great, it is said to increase without limit or to become infinite. The notation to represent this is x = », which is read "x increases without limit" or "x becomes infinite." lim The form „_ m [/(*)] = A is read "the limit of fix) as x becomes infinite is A" 169. Elementary theorems of limits. — The following theo- rems will be found useful in dealing with limits. They are given here without proof. • (1) If two variables thai approach limits are equal for all their successive values, their limits are equal. (2) The limit of the sum of a constant and a variable that approaches a limit is the sum of the constant and the limit of the variable. (3) The limit of the produdt of a constant and a variable that approaches a limit is the product of the constant and the limit of the variable. (4) If each of a finite number of variables approaches a limit, the limit of their sum is the sum of their respective limits. (5) If each of a finite number of variables approaches a limit, the limit of their product is the product of their respective limits. (6) // each of two variables approaches a limit, the limit of their quotient is the quotient of their limits, except when the limit of the divisor is zero. If the limit of the divisor is zero the limit of the quotient may have a definite finite value or the quotient may become infinite, but it is not determined by finding the quotient of Digitized by Google 222 ANALYTIC GEOMETRY [§170 the limits of the two variables. The calculus determines such limits as these exceptional cases. 170. Derivatives. — The fundamental conception of differ- ential calculus, and one that is of the greatest importance in mathematics, is the derivative of a function. Using the notation of this chapter the derivative is defined to be the At/ limit approached by the quotient ^ as Ax approaches zero. If the curve, Fig. 154, represents the function y = /0r), Ay the quotient ^- is the slope of the secant line PiP„ If Pi remains fixed and Ax approaches zero as a limit, the point P moves along the curve and approaches Pi as a limit, and the secant PiP turns about Pi to the limiting position QR, which is defined to be the tangent to the curve at the point Pi. Hence, the slope of (he tangent is precisely the quantity called the Fig. 154. derivative. It is evident that the value of the derivative depends upon the position of Pi on the curve. Definition. — The slope of a curve at any point is the slope of the tangent to the curve at that point. du The notation for the derivative is j-> read "the derivative of y with respect to x." Then by definition dy _ lim rAy~| dx~ Ax=0 LAxJ Of course, the independent variable and the function may du iim rAtn be represented by other letters. Thus, -37 = a*^o I aFJ _. . dy The notation -f- dx du is used to indicate the value of j- for x—xi ax the particular value X\ of x. /Google Digitized by* §171] ELEMENTS OF CALCULUS 223 Example. — Given y = x 2 , find dy and thus find the slope of the £-*+**• dx\x—2 1 tangent to the parabola at the point (2, 4). Also find the equation of this tangent and plot. Solution. — (1) Given y = x 2 . . (2) When x - 2, y - 4. (3) If a; takes an increment As, y + Ay = (2 + Ax) 2 = 4 + 4Ax + A?. (4) Subtracting (2) from (3), Ay - 4Ax + Ax 2 . (5) Dividing by Ax, (6) Letting Ax = 0, 5i|x = 2 ~ 4 * Hence the slope of the tangent to the para- bola at the point (2, 4) is 4. The equation of this tangent by [15] is y — 4 = 4(x — 2), or 4x — y = 4. The plotting is shown in Fig. 155. 171. Tangents and normals. — It fol- lows from the preceding article and [15] that the equation of the tangent to the curve y = f(x) at the point {x ly j/i) is [46] dy y - yi= di X-Xi (x - Xl ). +X Fig. 155. Definition. — The normal to a curve at any point is the line perpendicular to the tangent to the curve at that point. Then by [9] and [15] the equation of the normal to the curve V — f( x ) a * the point (x h j/i) is [47] . y-y 1 =-_l_(x-X 1 ). dy dxx=xi Example. — Find the equations of the tangent and normal to the ellipse 4x 2 + 9y 2 = 36 at the point (xi, y{). Also find these equations when xi = 2. Plot. Solution.— (1) Given 4x 2 + 9y 2 = 36. (2) Let x = xi and y = y u 4xi 2 + 9yi 2 = 36. If x takes the increment Ax, y will have the increment Ay, and ' (3) 4(xj + Ax) 2 + 9(2/i + Ay) 2 - 36, _ or 4xi* + 8xiAx + 4Ax* + W + lSyiAy + 9Ay* = 36. Digitized by Google 224 ANALYTIC GEOMETRY [§171 (4) Subtracting (2) from (3), 8siAa: + 4A3* + ISyiAy + 9Ay* - 0. (5) Transposing and arranging, ~ = Sx t +4As 18yi + 9Ay Passing to the limits and noticing that Ay * as Ax * 0, dyl __ 4a?i Substituting in [46], the equation of the tangent is y -Vi 4«i / \ or 4x& + 9yiy - 4*1* + 9yi 2 . Since by (2) 4»i f + 9yi f — 36 the equation of the tangent is 4*1* + 9y# - 36. Similarly the equation of the normal is y ~ Vl " te (X " Xl) ' When a; - 2, y - ±f V5- J\W4V») ^Jf (2.-^^6) Fig. 156. Substituting these values for xi and y\ in the equation of the tangent, the equation of the tangent at (2, f y/b) is 4-2x + 9-fVoV - 36, or 4s + 3\/% - 18. And the equation of the tangent at the point (2, — fv/5) is 4-2* +9(- W&)y =36, or 4s - 3\/% - 18. Likewise the equations of the normals are, by [47]: at the point (2, K/5), 9Vpx - 12y - 10\/5; and at the point (2, ~f a/5), 9\/6s + 12y = 10V5. The plotting is shown in Fig. 156. Digitized by Google §172) ELEMENTS OF CALCULUS 225 EXERCISES 1. Given y — x* t compute the values of Ay and ~ when x « 0.5 and Ax « 1, 0.1, 0.01, and 0.001 respectively. 2. Find the slope of the tangent and normal to y — 4x* at the point where z * 0.5. dy\ 3. Given y — x* + 2, find -p — « and write the equation of the tan- gent and normal at this point. Check the result by plotting. Find -£■ at the point G&i, yi) for each of the following: 4. y - 3s* - 1. 10. y - x* + 2s J + 5. 5. y ■ - x* + 4. 11. y - 3x* - 4x* + Ox. 6. y - 2x + 5. 12. x* + 2y» - 16. 7. xy - 4. 13. 4x* - 9y» - 36. 8. y* - 2px. 14. y* - 4x + 8. * 1 «. s + 1 16. Given the parabola y* — 2px, find the equations of the lines tangent to the parabola at the extremities of the latus rectum, and show that they meet on the directrix. 17* Find the slope of the circle x % + y* — 25 where x — 2, (a) when the point is in the first quadrant, and (6) when the point is in the fourth quadrant. 18. Find the angle that the line 3x — 4y + 7 » makes with the circle x* + y* — 25 at their point of intersection in the first quadrant. 19. At what angle does the circle x* + y* = 16 intersect the circle x f + y f = 8x at their point of intersection in the first quadrant? ALGEBRAIC FUNCTIONS 172. Differentiation by rules. — The process of finding the derivative of a function is called differentiation. The method used in the preceding articles in finding the derivative is? called the fundamental method since it is based directly upon the definition of a derivative. The derivative of any function can be found by this method, but the work can be greatly shortened by using rules or formulas which can be established by' fundamental methods or otherwise. The rules needed in differentiating algebraic functions will be 15 Digitized by Google 226 ANALYTIC GEOMETRY [§173 considered first, and later some of those necessary to differentiate trigonometric, exponential, and logarithmic functions. In the formulas, x, y, u, and v denote variables, which, of course, may be functions of variables, and a, c, and n denote constants. 173. The derivative when f (x) is x. — Since the equation y = x represents a straight line with slope equal to 1, and by dy article 170, -£ is the slope of the curve at any point, it follows that I <* = ! dx In general, the derivative of a variable with respect to itself is unity. 174. The derivative when f (x) is c. — Since y = c is the equa- tion of a straight line with slope equal to 0, it follows that n. £ = o. dx In general, the derivative of a constant is zero. 175. The derivative of the sum of functions. — Given y = u + v, where u and v are functions of x, and let Ay, Ati, and At; be the increments of y, u, and v } respectively, corre- sponding to the increment Ax. Let x = xi, then y\ = U\ + v\. Let x = Xi + Ax, then yi + Ay = u x + Aw + V\ + Av. Subtracting, Ay = Au + At;. Ay Au Av Dividing by As, aS = aS + A^' __ du i^HL X — Xi dX X = Xi dX X**X\ It is evident that any number of functions can be treated in a similar manner, then Let Ax = 0, then -/ ' dx ttt d(u + v + w+ • • -) _ du dv dw m * S " dx + dx "*" dx + Digitized by Google §176] ELEMENTS OF CALCULUS 227 Or, the derivative of the sum of any number of functions is equal to the sum of their derivatives. Example. — If y = x* + 3a;* — 4z + 3, dy = dM _ d(3s») _ d(4x) _ rf(3) dx dx dx dx dx 176. The Derivative of the product of two functions. — With the notation as in the previous article, given y = uv. Let x = Xi, then j/i = urn. Let x = Xi + Ax, then yi + Ay = (t*i + Aw)(vi + At;). Subtracting, Ay = wiAt; + v\bu + Aw At;. Dividing by Ax, Ay At; , Am , A At; a — ^i 7 V *>i 7 — hAw— • Ax Ax Ax Ax At; Let Ax = and notice that Au-^- also approaches zero as a limit, then IV. dy dx — dt; x=xi dx , du x=a;i dx X=*X\ d(uv) dv , du Or, #ie derivative of the product of two functions is equal to the first times the derivative of the second plus the second times the derivative of the first Example. — If y dy dx (x - 2)(s* + 1), S-b-t ^ + n + v + i)*^*!. dx dx 177. The derivative of the product of a constant and a function. — Given y = cu, where c is a constant. By the previous article But V. dy __ du . dc dx dx dx dc — - = 0. dx d(cu) du By II. Digitized by Google 228 ANALYTIC GEOMETRY [§178 Or, the derivative of the product of a constant and a function is equal to the constant times the derivative of the function. Examplea.^-Ify - 4(s - 2), ^ - 4 d(x ~ 2) - T - • u dy ldu "a dx adx 178. The derivative of the quotient of two functions.— Given u „-- Let x = xi, then yi = — • V\ Let x = xi + Ax, then yi + A# = l Subtracting, Ay = Vi + Av U\ + Am _ Wi _ yiAu — U\Av Vi + Av Vi "~ Vi(vi + Av) Aw At; Aj/ Ax Ax Dividing by Ax, ^ = ^^ ^ } du dy I r <ix dtl x*=xi ax \x**x\ «i» du dv V di dz Let Ax = 0, then , - , — ax \x=xi vi. -'•-¥-- dx v 2 Or, the derivative of the quotient of two functions is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominalor. r, i re x — 1 dy dx dx 179. The derivative of the power of a function. — Given y = u n . (a) When nis a positive integer. Writing as a product y = uu*~ l . r*. dy m .du , d(u H -*) ^ T - r - Digitized by Google §179] ELEMENTS OF CALCULUS 229 Writing w*" 1 as the product uu*-*, ax ax When this process is performed n times the last term will d(u n ~ n ) contain — -3 — -> which is zero by II. vn. .-. ^) = » u - l J 1 . dx dz (6) When n is a fraction, — , toAere p and q are positive integers. p Given y = u« Raising both sides of the equation to the qth power, y q = u*. Then M 9 " 1 ^ = vuV ~ l ~T' By ( a ) of this article - SI" f ^ ^ — P***"" 1 ^ g da;' dx " gj/*" 1 dx ©u^ 1 du p --1 du 5.(8-1) dx g dx qu« dx g dx (c) TTfcen n ts negative, either integral or fractional. Let n = — m. Then y = w- m = — Clearing of fractions, yu m = 1. Then myu^ 1 ^ + w- ^ = 0. By IV, VII, and II. Digitized by Google 230 ANALYTIC GEOMETRY [§180 Solving for & £ = - *2^ ** ax ax u m ax dx dx k Therefore formula VII is established when the exponent is a positive or negative integer or fraction. It is expressed in the following rule: The derivative of a function affected by an exponent n is equal to n times the function affected by the exponent n — 1, times the derivative of the function. Examples.- -If y - (** + x + 1)*, ^ = 4(s« + x + l)' ^*' + * + 1} « 180. Summary of formulas for algebraic functions. — The formulas here summarized enable one to differentiate algebraic functions. I - = 1 x * dx l * n. £ = o. dx m d(u + v + w H ) __ du dv dw U1# dx ~~ dx "*" dx "*" dx + ' ' % TTy d(uv) dv . du IV. \ ' = u r + v-j — dx dx dx V - dx " c dx VI. o du dv dx dx dx v 2 ._ d(u w ) n .du VII. -~- = nu*- 1 ^ — dx dx Digitized by Google §181] ELEMENTS OF CALCULUS 231 181. Examples of Differentiation. — If the formulas and rules of differentiation are well learned, their application is one of the easiest processes in mathematics. Example 1. — Given y = 7s 8 , find -p . | = «2g> = 7 ^g = 73x» | = 21*'. By V, VII, and I. dv Example 2. — Given y = x* + 2x % — 5s + 6, find -p. ds dx dx dx dx - 3s* + 4s - 5. By VII, V, II, and I. Example 3.— Given y - (s 2 + 2s) (3s - 2), find-^. g = (x , + !te) ^l2) +( 3 ;c _ 2) ^!^) By IV. - (s 2 + 2s)3 + (3s - 2)(2s + 2) - 9s* + 8s - 4. By I, II, III, V, VII. Example 4. — Given y ■» = — — r> find -X • (3,-1)^^ -(,.+2)^^ <*y_ ' *c dx B VI ds " (3s - l)i * y V1 - _ (3s - l)(2s) - (s 2 + 2)3 _ 3s 2 - 2s - 6 (3s - 1)* ~ (3s - 1)* Example 5. — Given y = ^x 2 + 3s, find ~. dy _ d^x* + 3s _ d(s 2 + 3s)* _ ... , d(s 2 + 3s) ds S di i{X +6X) di - *(*» + 3s)~*(2s + 3) .- -j^= • 3V (s* + 3s) 2 EXERCISES In the following find the derivative of the function with respect to the independent variable. 1. y - 3s 2 . 6. y = 4 Vs. 11. y = -17yV8. 2. y - 5s*. 7. y - 3-^s. 12. y = -2^/s 4 . 3. y = 7s*. 8. y - -^s"». 13. « - igt*. 4. y = as*. 9. y - 3s"*. 14. « = 4*.' 5. y = fs*. 10. y - -4s*. 16. 8 = JA 16. y =» s 4 + 3s 2 -f 2. 17. y - 3s 2 - 2s + 6. Digitized by Google 232 ANALYTIC GEOMETRY [§181 18. y - x« - x» + 3*. 22. y - (3s* + 2)« - 2s. 19. y - a* - ar* + 4. 28. y - (2x + 3)* - 3s. 20. y - x* - 3ar« + 2. 24. y - \/2x» - 7x. 21. y - (2x + 1)' - 3. 25. y - -^x« + 7x - 2. 26. „--,. 29. f-jqpj- 82. f-^jrj- 2x« Q * # * (x« - 1)* ™_*_ a; 8 + 1 85. y - 3a* - 4x • + 3a; 4 - 3. 48. « - VT+T + <^2* - 3. 86. y - Vx~+~1 - Va^T- 44. 8 - t* + 2T* + 3*«. 37. y - V3a; 8 + 7x* - 3a; + 2. 45. y - (a; 8 + l)(x* - 2a; + 1). 88. y - Vax* + bx +_c - Vx~+"S- 46. y « (x + a)»(x - 6)» 39. y - x»(x* + 5)*. 47. y - (a; + l)«(2x - 1)«. 2a; - 1 AO 2x* - 1 51. Find the slope of the tangent line to the curve y — a;* at the point where x = 0. At the point where x = 1. Where x ■» 2. 52. In exercise 51, what is the slope of the curve at each of the points? How many times faster is y increasing than x at each of the points? 53. If a point is moving from the origin along the curve y — 2x* in the first quadrant, what is the relative rate of increase of x and y when x = 1, 2, and 4, respectively? 54. Find the equations of the tangent and the normal to the curve y = x 9 + 4a; 1 + x - 6 at the point (0, - 6). At the point (2, 20). 55. In the curve of exercise 54, where is the tangent line parallel to the x-axis? 56. Find the equations of the tangent and the normal to the curve y = x + -j at the point (xi, yj. 57. Find the point on the curve y — x* + 3x* — 4x — 12 at which the tangent has a slope of — 7 V What is the equation of the tangent at this point? Plot the curve. 58. At what angle does the line y = x — 1 intersect the parabola y* + 4x - 4? Digitized by Google §182] ELEMENTS OF CALCULUS 233 x 9 59. Show that the parabola y* — Aax and the cissoid y — s — ^T~ intersect at right angles at the origin. 60. The heat H, required to raise a unit weight of water from Q°C. to a temperature f, is given by the formula H - t + 0.00002** + 0.0000003* 8 . JIT JTT Find -rr and compute the value of -=r- where t° — 35°C. 182. Differentiation of implicit functions. — In the previous exercises, the dependent variable in each was expressed as an explicit function of the independent variable. Often it is either not convenient or not possible to express one variable as an explicit function of the other. In such a case the usual 1 rules for finding the derivative can be applied and the desired derivative found as an implicit function of the variables involved. The method can be best illustrated by examples. Example 1. — Given x* + y % =» 25, find -p as an implicit function of x and y. Since y is a function of x, the left hand member is the sum of two functions of x. Differentiating, 2x + 2yj- =0. dy _ x dx y Example 2. — Find the equation of the tangent line to the curve x % — y h + x % — y « at the point (1, 1). Solution.— Differentiating, 5x 4 - 5y A -p + 3s 2 - ^ «0. Solving for -^t When x = 1 and y = 1, Then the slope of the tangent at (1, 1) is f . Hence the equation of the tangent is y — 1 — f (x — 1), oi 4x - 3y - 1 - 0. dy 5x* + 3z J dx dy dx s 5y* + 1 4 = 3. Digitized by Google 234 ANALYTIC GEOMETRY [§182 EXERCISES In the following find the derivatives as implicit functions. 1. s' + 2,« = ^findg. • 2. y* + y = s* +.*, find ^. S.*+g-l f fad*. a* 6 2 ' dx 4. pv = c, find -p and -r-« 5. x* - 4*V + y» = 0, find g| and p. 6.**+y* = o^, find g and g. 7. x 1 + y 1 - a 1 , find ^[. 8. (s + y)» + Or - y)» - a, findg. • t.(p+5)(.-»-MI-dJ M id* 10. Find the equations of the tangent and the normal to the circle x* + y % « 25 at the point (3, 4). 11. Find the equations of the tangent and the normal to the circle x* + y* - 4x + 6y - 24 = at the point (1, 3). 12. Find the equations of the tangent and normal to the ellipse 162 s + 25y* = 144 at the point in the first quadrant where x — 2. Show that the tangents to the following curves at the point (x h y t ) are as given. Equation of curve Equation o f tangent 13. z* + y* = r*. xis + y# = r*. H. y 2 = 2px. yiy = p(s + Xi). 15. a; 1 - 2py. XiX - p(y + yi). a 2 6 2 o f o f . a 2 6 2 = * 18. rry = c. x x y + yix - 2c. 19. Find the equations of the tangent and normal to the parabola 3* + y* = a* at the point (*i, f/i). a 2 6 2 Digitized by Google §183] ELEMENTS OF CALCULUS 235 FURTHER USES OF THE DERIVATIVE 183. Discussion. — By methods of analytic geometry the properties of the locus that are most conveniently discussed are the intercepts, symmetry, and extent (See Art 43). By means of the derivative other properties may be discussed. Some of these will be considered in the following articles. The discussion will be confined to equations (1) whose curves have no break, at least in the part of the curve con- sidered; and (2) where for each value of the independent variable there is but one point on the curve. Such curves, as well as the functions giving rise to them, are said to be continuous and single-valued. 184. Properties of a curve and its function. — If the curve, Fig. 157, is thought of as traced by a moving point passing from left to right, the following properties may be noted: (1) The curve is falling from A to B, from D to F, and from H to J; and the corresponding function is decreasing. (2) The curve is rising from B to D, from F to H, and from J to K; and the corresponding function is increasing. (3) If the curve rises to a certain position and then falls, such a position is called a maximum point of the curve. D and H are such points. The ordinate, that is, the value of the function, at such a point is called a maximum ordinate or maximum value of the function. (4) If the curve falls to a certain position and then rises, such a position is called a minimum point of the curve. B, F, find Digitized by Google 236 ANALYTIC GEOMETRY [§186 J are such points. The ordinate, that is, the value of the function, at such a point is called a minimum ordinate or a minimum value of the function. (5) The curve is concave upward between A and C, E and <?, and I and K. It is concave downward between C and E, and G and I. (6) Points C, E, G, and J where the concavity changes, are called points of inflection. Curves may have other peculiarities, but these will not be considered here. 186. Curves rising or falling, functions increasing or decreasing. — Since by definition, Art. 170, the slope of a curve at any point is the pame as the slope of the tangent at that point, it follows that when the slope is positive the curve is rising, and when the slope is negative the curve is falling. This is, of course, when passing from left to right. Stated with reference to the function this becomes the following very useful principle: When the derivative of a function is positive, (he function increases as the independent variable increases; when (he derivar- tive is negative, the function decreases as the independent variable increases. It also follows that the ratio of the change of the function at any point to that of the variable is equal to the value of the derivative of the function with respect to the variable, for that point. Example 1. — For what values of a? is the curve y «■ z* rising and for what values falling? Solution. — Given y — x*. Then %. - 2*. dx Now 2x is positive when x is positive, and negative when x is negative. Hence the curve is rising when x>0, and falling when x<0. Example 2. — For what values of 3 is the function y = x* increasing and for what values decreasing? Here -¥ = 3a; 2 , which is not negative for any value of x. Digitized by Google §186] ELEMENTS OF CALCULUS 237 Hence the function is never decreasing. Is it always increasing? Example 3. — For what values of 3 is the curve y — }*»— \x* — 6* rising and for what values falling? For what values of x is y increasing 6 times as fast as x? Solution, — Given y — }»»— \x* — Ox. Then -^ - x % - x - 6. Factoring, ^ - (x + 2)(* - 3). Then -p is positive when x< — 2 and when »>3, and negative when -2<s<3. Hence the curve is rising when x<—2 and when x>3, and falling when -2<a;<3. The values of x for which y is increasing 6 times as fast as x can be found by putting x* — x — 6 — 6, and solving for x. This gives x — 4 or —3. EXERCISES Passing from left to right, for what values of x are the loci of the following equations rising and for what values falling? 1. y = 3x — 6. 8. y - x* — x* — 2x. 2. y - 4a; 1 + l&r - 7. 9. y - s» - 2a* + 3 - 3. 3. y - VS. 10. y(l + a; 2 ) - a. 4. y« - 8s*. 11. y(a* - 1)* - s». 5. y - a* + 3. 12. 6y - 2s 8 - 3a* - 12s - 6. 6. xy - 15. 13. y « x 4 — 6s 2 + &c + 6. 7. y - *» - 9x. 14. y - (*« - 1)*. 15. In exercise 8, how many times as rapidly as x is y increasing when x - 10? When x - 3? When 3 - -1? When x - 0? 16. In exercise 9, for what values of a; is y increasing 7 times as rapidly as a;? For what values of x is y decreasing 4 times as rapidly as a; is increasing? 186. Maximum and minimum. — From the definitions of article 184, it is clear that if a curve is plotted in rectangular codrdinates, the curve is rising at nearby points on the left of a maximum point, and falling at nearby points on the right. For a minimum point the curve is falling for nearby, points on Digitized by Google 238 ANALYTIC GEOMETRY [§186 the left and rising on the right. The student can readily state this with reference to the function. It is evident that at a maximum point or a minimum point like those shown in Fig. 157, the tangent line is parallel to the x-axis, that is, its slope is zero. It follows that these points can be determined from the f miction as follows: (1) Equate -j- to zero and solve for x. (2) Determine whether j- is positive or negative for nearby points on the left and right. dv dv A point where -r- = is a maximum point if -r >0 for dy nearby points on the left and -r- < f or nearby points on the right. A point where -r- = is a minimum point if -t- < for dii nearby points on the left and -r- > f or nearby points on the right. It is distinctly understood that these tests determine only such points as are illustrated in Fig. 157. For cusp maxi- mum and minimum points as shown in Fig. 158, the tangent is perpendicular Fig. 158. to the 3>a j^ 8 an( j hence -p =" « . Example. — Determine the maximum and minimum points of the func- tion y ■» x s — 3s f + 4 and plot the curve. Solution. — Given y - x* — 3x f -f 4. ^ - 3s* - Ox - Zx{x - 2). ax . ' . ■¥ = f or x = 0, and x = 2. dx ' dv When x < but near 0> ~ > .0 and the curve is rising. Digitized by Google §187] ELEMENTS OF CALCULUS 239 When x > but near 0, -?■ < and the curve is falling. ax . * . the curve has a maximum point when x = 0. When x < 2 but near 2, -]*■'< and the curve is falling. ax When x > 2 but near 2, -p > and the curve is rising. . * . the curve has a minimum point when x — 2. Plotting.— When 3 - 0, y = 4. . * . (0, 4) is a maximum point. When x = 2, y = 0. . * . (2, 0) is a minimum point. Factoring, y = (a + 1)(* - 2)(s - 2). . * . the z-intercepts are —1, 2, and 2. ,A few other points will make the plotting fairly accurate. See Fig. 159. X 1 3 4 - 2 V 2 4 20 -16 Fig. 159. EXERCISES Determine the maximum and minimum points of the following curves and plot. 1. y - x*. 5. y - (x + 4)(s - 2)(* - 4). 2. 2y - a 1 - 4* + 6. 6. y = s« - 7s* + 36. 3. y » 6x - s« + 4. 7. 16y = a; 1 - 32x. 4. 4s* + 9y* = 36. 8. y = s« - 4s«. 9. By finding the maximum point of the curve, find the coordinates of the vertex of the parabola 2x* — l&r + 15y — 21 — 0. 10. The equation of the path of a projectile is y = tan a-x — ; Art 94.) 2v* cos* a ' Find the maximum height to which the projectile rises. 187. Concavity and points of inflection. — It is evident from an inspection of a curve that is concave upward that the tangent line turns counter-clockwise in passing along a curve from left to right, that is, the slope of the tangent increases. Likewise, if the curve is concave downward, the tangent line turns clockwise, that is, the slope of the tangent is decreasing. Digitized by Google 240 ANALYTIC GEOMETRY [§187 Thus, in Fig. 160, the tangent line turns counter-clockwise in passing from A to Z>, and the slope increases from a negative value at A to a positive value at D. likewise, in Fig. 161, the tangent turns clockwise in passing from A to D, and the slope decreases from a positive, value at A to a negative value at D. Fig. 160. Fig. 161. It remains to determine how the concavity of a curve can be determined from its function. Since the derivative of a function of a: is itself a function of x, it is evident that the derivative of this first derivative may be found. It is called the second derivative of y with respect to x. If y = f(x) y the second derivative is jzviz) an d is d*y represented by the symbol -ry Thus, if y - x* — 6x* + \2x - 3. ^ - 3s« - 12* + 12, and p{ - 6s - 12. dx* d*y From the foregoing, it is evident that when -r\ is positive, -i| is increasing; and when -r- t is negative, -j- is decreasing. Or, if y = f(x) is the equation of a curve, the slope of the tangent is increasing when passing from left to right and the dhi curve is concave upward for the values of x that make -t4 positive. Digitized by Google . §187] ELEMENTS OF CALCULUS 241 Likewise, the curve is concave downward when -j-? is ' ax 2 negative. From (6) of article 184, it is evident that a point of inflection is a point on a curve at which the concavity changes from upward to downward or vice versa. A point of inflection can be determined by finding the values of x for which -r\ changes sign, providing the function is finite for that value of x. Example. — Investigate y ■» x* — 3x* + x + 2 for concavity and points of inflection. Solution. — Given y « x* — 3s* -f x + 2. P - &f - 6* + 1. g-te-6-6(*-l). Since when »<1, 6(a; — 1) is negative; and when «>1, 6(a; — 1) is positive, the curve is concave downward at the left of x = 1, and concave upward at the right of & = 1. Therefore, it has a point of inflection at the point (1, 1). EXERCISES In exercises 1-10 investigate for concavity and points of inflection. y - (a; + 2)(* - 2)(x - 3). y = 3s 4 - 4s 8 - 1. y = x* — 4a; 2 + 4x — 1. y = a; 4 - 2a; 2 + 40. y - 3a* - 16a; 8 - 6a; 8 + 48a; + 17. 11. In the example, Art 187, find the slope of the tangent to the curve at the point of inflection, find the maximum and minimum points, and plot the curve. 12. In the example referred to in exercise 11, if the curve is being traced by a point moving from left to right, for what values of x does y increase at the same rate as a;? How rapidly is the curve rising when x - 3 if x is increasing at the rate of 2 inches per second? 13. Investigate the greatest possible number of points of inflection of the curves of (1) y = ax 1 + bx -f c, (2) y = ax* + bx % + ex + d. (3) y - ax 4 + bx* + ex 1 + dx + e. 16 1. y — a; 8 . 6. 2. y - x 4 . 7. 3. y « a* 8. 4. y = 3a; - -a; 8 . 9. 5. y « x A - -6a;». 10. Digitized by Google 242 ANALYTIC GEOMETRY Hiss In exercises 14-19 plot the curves showing the values of y, -& and -z-^f using the same set of axes for the three curves of each. What facts can be read from these curves? 14. y - 4s«. 17. y - (x + 2)(x - 2)(x - 3). 15. y - 3x*. 18. y - x* - 12s + 7. 16. y - 3x - x 9 . 19. y = *r 4 - 2x* - 8. DIFFERENTIALS 188. Relations between increments. — When two variables are so related that the ratio of their corresponding increments is constant, either variable is said to change uniformly with respect to the other. When the variables are related by an equation of the first At/ degree, as y = mx + 6, where Ay = mAx, then ^- = m. That is, either variable changes uniformly with respect to the other. N M AXM- +-X AA C AXM Fig. 162. Fig. 163. This is also evident from Fig. 162, in which y : = mx + b is the equation of the line PP\ with slope m. P is any point op At/ this line and -r-^ = m. A3 In Fig. 163, BCDE is a rectangle having a constant altitude a and a variable base x. When a; takes an increment Ax, the area A will take an increment aAx. .\ AA = aAx When two variables are so related that the ratio of their AA or -r— = a. Ax Digitized by Google §189] ELEMENTS OF CALCULUS 243 corresponding increments is variable, either variable is said to change non-uniformly with respect to the other. If the variables s and t are related by the equation s = \gt 2 } then As = \g(2tAt + At 2 ). See Art. 167, example 2. As Here ^r is a variable for it varies with t, that is, differ- As ent values of t give different values of tt, and the change is non-uniform. 189. Differentials. — If two variables are so related that one is dependent and the other is independent, then for corresponding values of the variables: (1) The differential of the independent variable is the value of its increment. (2) . The differential of the dependent variable is what would be its increment, if at the corresponding values considered, its change became and remained uniform with respect to the independent variable. The differential of a variable is denoted by writing d before it. Thus, differential x is denoted by dx. Also dy, d(x*) t d(x 2 + 2x + 1), and df(x) denote the differentials of y, x* } x 2 + 2x + 1, and/(x), respectively. 190. Illustrations. — It follows from the definitions that the differentials of variables that change uniformly with respect to each other, are their corresponding increments. Thus, if y = mx + b, dx = Ax and dy = Ay, for y changes uniformly with respect to x. It should be noted that dy = Ay when, and only when, the graph of y = f{x) is a straight line. If the rectangle of constant altitude, Fig. 163, is increased in area by increasing the base by the length CM , the area is -increased by the rectangle CMND. Here evidently the area, Digitized by Google 244 ANALYTIC GEOMETRY [§190 A, is a function of the base x. Since A and x change uniformly with respect to each other, CM = dx and the rectangle CMND - dA. Consider the curve y = /(x), Fig. 164, as being traced by a point starting from the origin and moving to the right and upward. The direction that the tracing point is moving at any point is along the tangent line at that point. Let (x, y) be the coordinates of the moving point. Evidently, y is changing non- uniformly with respect to x. Suppose the moving point has reached Pi. Here y is evidently Fig. 164. changing at the same rate it would if the point were moving along the tangent line at P x . If then the change in y is to become and remain uniform with respect to x, the point must move along the tangent. It follows that at the point Pi, if the increment of x is Ax = MiM, dx = Ax, and dy = QT. It is to be noted that the corresponding increment of y is Ay = <2P. dy Further, if the slope of the tangent, -r-, that is, the deriva- tive, is represented by/'(x), dy = f (x)dx. Since dy and dx are finite quantities, dividing by dx, This is an extremely important and useful relation, for it states that the derivative and the ratio of the differentials can be used interchangeably. Again, referring to Fig. 164, if « is the length of the curve Digitized by VaOO 1 §190] ELEMENTS OF CALCULUS 245 traced, then corresponding to dx and dy, the change in s, if this change becomes and remains uniform, is ds = PiT, and ds 2 = dx 2 + dy 2 . The triangle PiQT is called the differential triangle. Example. — A point is moving along the parabola y = 3x*. When it has reached the point whose abscissa is 2, find dy and ds corresponding to dx = 0.1. Solution. — First find the derivative of y with respect to x. Given equation y = 3x*. dx ;. dy = 6s -dx, for any value of x. When x - 2 and dx * 0.1, d y = 6-2-0.1 - 1.2. And da - V<te* + dy 2 - \/0.1 2 + 1.2* - 1.2042—. EXERCISES 1. The right triangle, Fig. 165, is being generated by the altitude moving uniformly to the right. If the variable base is x and the area A, show that dA corresponding to dx is the rectangle M\MQP\. 2. The area of the upper half of the area of the parabola y* — 4x is being generated by the ordinate moving toward the right. If A is the variable area, show that dA — 2\/x dx for any value of x. Draw the figure. 3. If the upper half of the area of the circle x s + y* = r* is being generated by the ordinate moving uniformly toward the right, show that dA =* y/r* — x* dx. 4. The area above the s-axis between y « sin x and the s-axis is being generated by its ordinate. Show that dA = sin x dx. For the part below the z-axis show that dA = —sin x dx. 5. A point is moving on the circle x* + y* = 25. Find dy and d* corresponding to a change in x of dx = 0.2 at the point in the first quadrant where x » 3. 6. A point is moving on the ellipse 52 +ii ■ !• Find dy correspond- ing to dx = 0.4 at the point in the first quadrant where x — 2. In the second quadrant where 3 « — 2. Digitized by Google 246 ANALYTIC GEOMETRY [§191 In the following find dy for any x. 7. y - 3s 1 + 2x - 1. 11. y = Vx* + 4. 8. y = x 8 + 4x + 2. 12. x^ + ^ - 4. 9. y » s« - 3x» + 2x*. 19. x* + ^ - 2. m* £_ L 14 . y ,1 16 9 * VSm 1 ^ 15. The distance 8 that a body will fall in t seconds is given by the formula * = \qt*. Find d« for any value of t. Find ds when t — 2 and corresponding to (ft = 1. (Use g = 32.) INTEGRATION 191. The inverse of differentiation. — Just as division is the operation that is the inverse of multiplication, and the extraction of a root is the inverse of raising to a power, so differentiation has its inverse operation. Here, as usual, the inverse operation is the more difficult. In fact, it is frequently- impossible to do the inverse of a differentiation except approximately. The process of doing the inverse of a differentiation is called integration. The result obtained is called an integral. The methods of integrating can be dealt with here to only a very limited extent. In general, an integral is found by knowledge acquired from differentiation, by reversing the rules of differentiation, or by reference to a table of integrals. Integration has very many applications to problems arising in the sciences and in engineering as well as to problems in mathematics. The symbol, ,/*, indicates that the differential before which it is written is to be integrated. Thus, J*2xdx indicates that a function of £ is to be found whose differential is 2xdx. The function is evidently z* + C % where C is any constant, for <*(*» + O - d(x \ + Q dx - 2xdx. ax Since the differential of any constant is zero, the function sought when integrating may contain a constant no indication of which appears in the given differential. For this reason Digitized by Google §192] ELEMENTS OF CALCULUS 247 the integral of a different'al is, in general, indefinite, and is called an indefinite integral. The constant C that is supplied when integrating is called the constant of integration. 192. Determination of the constant of integration. — The constant of integration is determined by having some fact about the function given besides its differential. This can be best illustrated by examples. Example 1. — Find the equation of a curve such that the slope of its tangent line at any point shall equal to the abscissa of the point if, further, it is given that the curve passes through the point (2, 4). dy Solution. — Since — = slope of tangent, and dx x = abscissa of point of tangency, dy dx dy Considering — as the ratio of dy to dx, and multiplying by dx f dx dy = xdx Then J*dy = fxdx, and y = ix* + C. Here C is any constant, and the equation represents all parabolas having their axes on the y-axis and opening upward. Some of these are represented in Fig. 166. It is evident that one such parabola can pass through any particular point of the plane. The one sought passes through (2, 4), and there- fore these values must satisfy the equation y » ix* + C. Substituting (2, 4) in this equation, 4 = J.2* + C. :. C - 2. The equation of the curve satisfying both conditions is then y - iz* + 2. Example 2. — Find the area enclosed by the parabola y* =* 4x and the double ordinate corresponding to x = 8. Solution. — The parabola y 2 = 4x is shown in Fig. 167, and is sym- metrical with respect to the x-axis. Then one-half of the area is above the x-axis and is the area OMC. x. Fio. 166. Digitized by Google 248 ANALYTIC GEOMETRY [5193 H-I Consider the area A as generated by an ordinate moving from the origin toward the right. When it has advanced any distance x dA — y dx. But y = +2y/x since y is positive. Then dA - 2y/x dx. Integrating. A - fr* + C. (1) A further fact about the area A is that A — 0. when x « Substituting these values in (1) gives 0-0 + C. .\ C - 0. Hence for any value of x t A « \x* + 0. Andforx=8, A - l-8*=yV2 =30.17-. . * . the total area = 2A — 00.34— square units. 193. Methods of integrating. — While a knowledge of differentiation enables one to write at once the in- tegrals of many differentials, the following formulas will help in inte- Fio. 167. grating forms that occur frequently. (i) yvdu - iL-j + c Here u may be any function of which du is the differential, and n is not equal to — 1. That (1) is true can be readily proved by finding the differ- ential of — r— r + C. n -jr 1 Example 1. — Find fx*dx. Here x = u, dx — du t and n — 4. . * . ys 4 ^ =» fr 1 + C. Example 2.— Find f(x* + x*)*(3x* + 2a0<k. Here x* + a 1 - w, (3a; 1 -f 2s)<fo — du, and n — 2. . * . y (*» + x*)*(3a; J + 2s)d s - ./Vdt* - Ju»+ C - *(*« + *»)• + C. Example 3.— Find jf Vs* - 1 2xdx. Here a* — 1 = w, 2a;da; = du, and n = J. .-. yVa^H 2a**r - yw*dw - |w* + C - §(*« - 1)1 + C. (2) /cdu = c/du. Digitized by Google §193] ELEMENTS OF CALCULUS 249 This states that a constant can be written either before or ' after a sign bf integration. Proof. — Since d(cu) = cdu. By differentiation. Then cu = J* cdu. By definition of an integral. But fdu = u. By (1) where n = 0. And cfdu = cu. Equating values of cu .'. J* cdu = cfdu. Example.— fbx*dx - bfx*dx - 6 • \x* + C - fa: 4 + C. (3) y (du + dv) - S&vl + f&v. Proof. — d(u + v) = du + dr. By differentiation. Then y*(dw + dv) = w + v. By definition of an integral. But u = fdu and t; = y\fo. By (1). . * . f (du + dv) = y*dw + fdv. This can readily be extended to the integral of the sum of any number of differentials. Example.— f(x* + 3s* - x + l)ds = fx*dx + y3aj*da; - yxcfo; + fdx By (3). -' \x* + a' - fcr 2 + a + C. By (1) and (2). Here C is the sum of the several constants of integration. EXERCISES Find the indefinite integrals in exercises 1-10, and check by differ- entiation. 1. dy = 4xdx. 6. dy - (2x + l)dx. 2. dy « x*dx. 7. dy = (2a; 2 + x + 2)da\ 3. dy « 4a;*da;. 8. dy - (a; — l)(x + l)ds. 4. dy - aAto. 9. dy - (a; + l) 8 da;. 5. dy « a^ds. 10. dy = (a + l)*da\ 11. Find the equation of the curve whose slope at any point is equal to three times the abscissa of that point, and which passes through the point (2, 6). 12. Find the equation of the curve whose slope at any point is equal to the square of its abscissa at that point, and which passes through the point (1, 1). 13. Find the equation of the curve whose slope at any point is equal to the square root of its abscissa at that point, and which passes through the point (2, 4). Digitized by Google 250 ANALYTIC GEOMETRY [§194 14. Find the area enclosed by the parabola y 1 — 2x and the double ordinate corresponding to x * 4. 15. Find the area enclosed by the parabola y* — 3x, the z-axis, and the ordinate corresponding to a: — 2 and x = 8. 16. Find the area between the curve y — 2a& and the z-axis from the origin to the ordinate corresponding to x = 10. Check by finding the area considered as a triangle. 17. Find the area between the curve y - x x and the z-axis from the origin to the ordinate corresponding to x '*» 4. 18. Find the area between the curve y — x % and the z-axis from the ordinate corresponding to x = —3 to the origin. 19. Find the area enclosed by the semi-cubical parabola y = z* and the double ordinate corresponding to x — 4. 20. Find the area enclosed by the curve y — z""*, the z-axis, and the ordinates corresponding to x = J and z = 8. 21. Find the area that is below the z-axis and is enclosed by the parabola y — z* — 4z + 3 and the z-axis. 22. Find the area that is below the z-axis and is enclosed by the curve y = z» — 4z* + 3z and the z-axis. TRIGONOMETRIC FUNCTIONS 194. So far in the calculus a study has been made of alge- braic functions only. The trigonometric functions will now be considered to a limited ex- tent. The sine and cosine will receive the chief attention, the formulas of the others will be given for completeness only. 195. Derivatives of sin u and cos u. — Let be a unit circle generated by the point P (x, y) moving in the positive direction, Fig. 168. Let u be the measure of the angle XOP in radians, and let s be the measure of the arc XP in linear units. Then u = s, x = cos u } and y = sin u. Differentiating, du = ds, dx = d(cos u), and dy = d(sin u). Fig. 168. Digitized by Google §195] ELEMENTS OF CALCULUS 251 In the differential triangle PQT, dx = PQ, dy = QT, and ds = PT. Also the angle at T through which the tangent has turned is equal to u. Now dy = ds* cost*, cosw and dy having the same sign. But dy = d(sin u) and ds = du. . *. d(sinw) = cos u du. Dividing by dx gives the derivative formula: ttttt d(sia u) du VIH. -~i — - = cos u -j— dx dx Also dx = —ds sin u, sin u and dx having opposite signs. But dx = d(cos u) and ds = du. .'. d(cosu) = — sin udu. Dividing by dx gives the derivative formula: _ d(cos n) . du K. _L_J = _ smu _. It is to be noted that the derivation of VIII and IX requires that the angle shall be in radians. Example 1. — Given y = sin (3x* + 4x — 1), find ~ Solution.- dg = d8in(3*»+4*-l) _ dx ax = cos (8** + te - 1) d(3 *' ^ " 1} - By VIII. h ,^+fc-D ii W) jW.fl t <ix dx dx ax ^ - (6x + 4) cos (3x* + 4x - 1). ox Example 2. — Find the maximum and minimum points of the curve y — cos x. Solution. — -]r — — sin x. dx Putting -p = gives — sin x ■» 0. x = nx, where n = 0, ±1, ±2, • • •. For values of x near an even number of times x but less than x, — sin x is positive; and for values of x near an even number of times x, but Digitized by Google 252 ANALYTIC GEOMETRY [§196 greater than *-, — sin x is negative. Hence the curve is rising before and falling after x - 2nr. . * . maximum points are the points for which x =• 2nx. likewise minimum points are the points for which x — (2n + l)x. Example 3. — Find the area enclosed by an arch of the curve y * sin x and the z-axis. Solution. — The curve y « sin x is shown in Fig. 169. The area sought extends from r^JT x * to x — r. Consider this area A as being generated Fio. 169. k y ^ ordinate moving toward the right. Then dA — yds - sin x dx. And y dA - y*sin x dx. . * . A — — cos x + C. By the inverse of differentiation. When x - 0, A =» 0. . • . « -cos + C, or C - 1. When a? « r, A = —cos x + 1 = 2 = number of square units in area. 196. Derivatives of other trigonometric functions. — The following formulas are stated for completeness. Their deriva- tion is not difficult and may be performed as exercises. _ d(tan u) 5 du X. — *-j — - = sec 2 u -=-• dx dx _ d(cot u) , du XI. -*-= — - = -csc 2 u^— dx dx _. d(secu) . du XIL — —i — - = sec u tan u -=-• dx dx __ d(csc u) . du Xlil. — ^-j — - = —esc u cot u j— dx dx _ t7 d(vers u) . du XIV. — ^ = sin u -i— dx dx XV. XVI. du d(sin~ 1 u) _ dx dx "~ VI - u 2 du dCcos- 1 u) dx dx Vl - u 2 /Google Digitized by* §197] ELEMENTS OF CALCULUS 253 xvn. xvra. XIX. xx. XXI. du dCtaii- 1 u) dx dx 1 + u* du d(cot-* u) dx dx 1 + u* du d(sec- x u) _ dx dx " uVu 2 - 1* du dfosc-^u) _ ^ dx du d(vers _1 u) dx dx \/2u — u 2 197. y*sin u du and y*cos u du. — Many integrals involv- ing trigonometric functions occur in the applications of cal- culus, but here attention will be confined to y*sin u du and y*cos u du. y* sin u du = — cos u + C. This is readily proved, for d(— cos u + C) = d( — cos u) = sin u du. y*cos u du = sin u + C For d(sin u + C) = d(sin u) = cos udu. Example. — Find J*Bin(2x + l)dx. If 2x + 1 - u, du = 2 dx. Then write ./"sin (2x + l)dx in the form lj*8in(2x -f l)2dx and it is in the form of J* sin u du. .\/sin(2s + l)<& - ifBm(2x + l)2dx = -Jcos (2s + 1) +C EXERCISES In exercises 1-20 find the derivatives. 1. y ■» sin 3a;. 5. y - sin 3s cos 2x. 2. y - sin 2 x. 6. y - tan 3s. 8. y - cos (2x H- 1). _ sin a? 4. y - sin * cos x. • V m C08 x " Digitized by Google 15. V « sin* x\/sec x. 16. V = m cot n qx. 1 — cos X 17. V ~ 1 + cos x' 18. P - tan 30 + sec Z$. 19. P - J taD s - tan + $. 20. V = 3 sin x — 4 sin 1 x. =» a (1 — cos 6), find dx and dy, then 264 ANALYTIC GEOMETRY [§198 8. y - tan 1 5x. 9. y — 3 sin x. 10. y - sin (x» + x*). 11. y « sin (x* + Zx - 4). 12. y - co s 2 (3s + 2). 18. y - Vsin 3x. 14. y = sin'x cos*x. 21. Given x = a(0 — sin 0) and y by division find -p • 22. Find the area enclosed by one arch of the curve y = cos x and the x-axis. 28. Find the slope of the tangent to the curve y — sin x at the point where x =■ Jr. Where x — 2. 24. Find the slope of the tangent to the cycloid x — a(0 — sin 0), y — a(\ — cos 6) at the point where 6 = ix. Where = *■. Where =0. 25. Find the maximum and minimum points, and the points of inflec- tion of the curve y — sin x. Find the indefinite integrals in exercises 26-33. 26. f sin 3x dx. 80. jT sin x cos x dx. 27. y* sin (3x — l)dx. 81. y* sin 1 x cos x dx. 28. f cos 4x dx. 82. y* cos* x sin x dx. 29. y* cos (4x — 2)dx. 88. y* sin* x cos x dx. 34. Find the equation of the curve passing through the point (x, 0), if the slope of the tangent at any point is equal to the cosine of the abscissa of that point. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 198. Derivative of log e u. — We will first find -?- when y = logeX.* Let P(x h t/i) be a point on the curve. Then j/i = logcXi or Xi = & l . Let x = Xi + Ax y and X\ + Ax = e vl+ **. Subtracting, Ax = & l+ *» - e vi = e vi (e* v - iy. *In log,x, e = 2.71828 • • •, the base of the natural system, of logarithms. .:..-■ Digitized by Google §190 ELEMENTS OF CALCULUS 265 Dividing by Ay, — = • 1 10 — = Xi • Ay Ay Ay Or ^L « L A ^ when Ax = 0. or ^= i iim r Ay 1 dx xi Ay = Le^ - 1J But it can be shown that A lu ? A [ — Ay , 1 - 1. Ay = Le** — 1 J Then, dropping the subscripts, •¥- = -, or dy =- dx> cfx X X Evidently, if 2/ = log«u, then dy = - dw. Dividing by <fc,g = i g- d(log«u) _ 1 du dx u dz 199. Derivative of log A u. — Let a be any base. Since by a theorem of logarithms, log u = log«vlog a e. Then <*^ = <*0^i ogo e. ByV XXm. .-. *flg*>-Jgloi*. By XXII. If, in XXIII, a = 10, logi w expresses the common logarithm of u, and djlogiou) _ 1 dit , , dx ~~ u dx ' where ilf = logioe = 0.4343 — . 200. Derivative of a tt and e tt .— Let y = a". Then log e y = u log«a. Digitized by Google 256 ANALYTIC GEOMETRY [§201 Taking the derivative of each side of this equation by XXII and V, 1 dy du* ~ dy du 7 «»• •■ ^ -*£«** If o is put equal to e, and noting that log«$ = 1, XXIV becomes XXV. ^=e«£- dx dx 201. Derivative of u\ — Let y = u v , where u and v are functions of x. Then log e j/ = v log«w. Taking the derivative of each side of this equation by XXII and IV, ldy^vdudv. y dx" u dx dx ** ' dy v du , dv , Tx = * uTx+ y te Xo *< u m v du . „ cfo . The application of formulas VII, XXIV, and XXVI should be carefully distinguished. Formula VII is used when a variable is affected by a constant exponent; XXIV is used when a constant is affected by a variable exponent; and XXVI is used when a variable is affected by a variable exponent. It is customary in calculus to omit the base when writing logarithms to the base e, and to express the base when it is not e. Thus, log 5 means log*5. Digitized by Google §202] ELEMENTS OP CALCULUS 257 202. Illustrative Examples. Example 1. — Given y = log (X* + 3x), find -=-• dy __ dx ~~ 1 d(x x» + 3x 2 + 3x) f da? (9-r xt + Sx U* , «,. . <fy a 2x +3 " da? = x» + 3x By XXII. By III, VII, V, I. Example 2.— Given y - logi (l + 3s), find -p. Tx - r+E - s — log10 *' By XXIIL 1 31og 10 e. By III, V, I. l + 3x . dy 3 * • " dx = 1 + 3x logio«. Example 3.— Given y - e* ,+x , find & p_^*d(£+z) t xx - e* ,+ * (2x + 1). By III, VII, I. Example 4. — Given y = log sin*x, find ™ dy 1 d(sin'x) dx sin*x dx By XXII. - gjig 2 sin x cos x. By VII, VIII, I. .'.^ = 2cotx. dx Example 5. — Given the catenary y = \a{eP + e a ), (see exercise 8, page 167), find the slope of the curve at the point whose abscissa is 0. 17 Digitized by Google 258 ANALYTIC- GEOMETRY [§203 Solution.— Given y - \a{e a + e °). By HI, XXV. By V, I. . dy • 'dx dy\ J(e° - e °). ax\x = U 203. the slope of the catenary at the point where x = is 0. J. I — f yVdu, and yVdu. — These integrals are readily evaluated and occur frequently. u For d(log.u + C) = — • By XXII. u £ = lOgeU + C. ++»x Fig. 170. y*e tt du = e u + C. For d(e u + C) = e tt du. /aMu - t -^- + C. For d(r^— + C) = a tt du. Mog a / log a By XXV. By XXIV. Digitized by Google §203) ELEMENTS OF CALCULUS 259 Example. — Find the area bounded by the equilateral hyperbola xy = 4, the x-axis, and the ordinates corresponding to x = 1 and x = 8. Solution. — The hyperbola is shown in Fig. 170, and the area sought is MNQR. Consider the area A as generated by the ordinate moving toward the right. Then dA = ydx. And fdA - fydx - f*dx - ±§~ A - 4 log x + C. When re - 1, A - 0. . ' . - 4 log 1 + C, or C - -4 log 1. When x = 8, A = 4 log 8 - 4 log 1. A - 4 X 2.079 -4X0 = 8.316. Therefore the area sought is 8.316 square units. EXERCISES In exercises 1-20 find the derivatives of the dependent variables with respect to the independent variables, and the differentials of the dependent variables. 1. y = log (x 2 + 7x). 7. y = e 8 **". 2. y = logio x*. 8. y = e* sin x. 3. y = log -• 9. y = a 2 *. X 4. y = logio or*. 10. y = xl0 2 * + *. 5. y - e 2 *. 11. y - (3s - 2)*. 6. y = e**. 12. y - J(« x + e"*). 13. i = 6e-°«. (See Ex. 9, page 167.) 14. i = /e L. (See Ex. 11, page 167.) 15. y — e~* sin x. (See Ex. 18, page 174.) 16. i - e~*' sin (2* + Jr). (See Ex. 19, page 174.) 17. y = * + log (1 + x 2 ). 19. y = (3x + 2)<r* 2 . 18. y - (2x + log x) 2 . 20. y = ( x * + 1) 2 *+ 8 . 21. Find the slope of the tangent to the curve y = e* at the point where x = 0. Where x — 2. 22. Find the slope of the tangent to the curve y — logio x at the point where x — 1. Where re = 10. 23. Find the minimum point of the curve y = log (x 2 — 2x + 3). 24. Find the maximum and minimum points of the curve whose equa- tion is y = 2x % — log x. 25. Show that the rate of change of y with respect to x for any point on the curve y = oe* x is proportional to y. (See Art. 128.) Digitized by Google 260 ANALYTIC GEOMETRY [§203 26. Find the area bounded by the equilateral hyperbola xy — 1, the x-axis, and the ordinates corresponding to x - 1 and re — 10. 27, Find the area bounded by the curve y ■* x + -> the x-axis, and X the ordinates corresponding - x — 2 and x - 4. Find the indefinite integrals in exercises 28 to 37. -dx. 34. y(l - *-i)(l - x"*)cix. SO. J^-tI«te. 85. y\»*<te. 81 fco8*<te J SIT ^±^<te. 36. y(e* + 4)e-*d*. SUI X 32, fe**dx. 37. y (e«* +1 + x)dx. 38. Find the equation of the curve passing through the point (0, 1) if the slope of any point of the curve is proportional to the ordinate of the of that point. Suggestion.— ~r " ky. .'. — = kdx. ax y 38. Find the equation of the curve passing through the point (0, 1) if the slope of any point of the curve is equal to xy. Suggestion.— -p — xy. .' . — = xdx. ox y Digitized by Google CHAPTER XIII SOLID ANALYTIC GEOMETRY 204. Introduction. — In plane analytic geometry, all the points and lines are confined to one plane. In solid analytic geometry this restriction 'is removed, points and lines are considered as anywhere in space. In addition a new element is introduced, a surface of which the plane is a particular instance. Since plane analytic geometry is a special case of solid analytic geometry, it is expected that the formulas obtained for plane analytic geometry can be obtained as special cases of the formulas for solid analytic geometry. Such reductions and resemblances should be constantly sought. 205. Rectangular coordinates in space. — If at the origin of the coor- dinate system in plane analytic geometry a line is erected perpendic- ular to the plane of the axes, this line will serve as a third axis for a space coordinate system, and is called the z-axis. It is customary in a space depiction to draw the x-axis, Fig. 171, horizontal, the 3-axis vertical and the #-axis as coming toward the observer. In order to give space perspective to the figure, the positive y-axis is drawn so as to make an angle of 135° with the posi- tive z-axis, and the unit on the j/-axis is taken equal to half the diagonal of a square whose side is a unit on the ic-axis. The three axes determine three coordinate planes, the 261 i z V B / y O II M A / /" ? i £ Fig. 171. Digitized by Google 262 ANALYTIC GEOMETRY [§205 xy-plane, zz-plane and yz-plane. These three coordinate planes are mutually perpendicular to each other and all pass through the origin 0. If through a point P in space, Fig. 171, planes are drawn perpendicular to the x, y, and z-axes, respectively, these three planes will form with the coordinate planes a rectangular parallelepiped. The edge RP perpendicular to the yz-plane and parallel to the x-axis is called the x coordinate of P. It is considered positive if measured to the right, and negative if measured to the left. The edge NP perpendicular to the xz-plane and parallel to the y-axis is called the y coordinate of P. It is considered positive if measured toward the observer, and negative if measured away from the observer. The edge KP perpendicular to the xy-plane and parallel to the z-axis is called the z coordinate of P. It is considered positive if measured upward and negative if measured downward. These three coordinate lines uniquely determine a point P, since they determine three mutually perpendicular planes, MP, HP 9 and LP which intersect in one point P. In place of drawing a rectangular parallelepiped to rep- resent a point in space it is customary to draw a broken line consisting of three of its edges. Thus, the point P, Fig. 171, would be represented by the broken line OHKP. The three coordinates of a point are written (x, y, z.) Thus if P, Fig. 171, is the point (2, 3, 4), its coordinates are x - OH - 2, y = HK - 3, and z = KP - 4. The three coordinate planes divide all space into eight octants. The octant in which the point lies is denoted by the sequence of signs for the three coordinates. Thus, the (+, -f-, — ) octant is the octant to the right of the 2/z-plane, in front of the sa-plane, and below the sy-plane. Digitized by Google $206] SOLID ANALYTIC GEOMETRY 263 EXERCISES 1. If P, in Fig. 171, has the coordinates (2, 3, 4) what are the coordi- nates of H, K, M, R, L and JV? 2. Plot the points (1, 1, 1), (-1, 2, 3), (2, -3, 1), (-2, -1, -3). 3. Draw the triangle whose vertices have the coordinates (2, 1, 4), (-1,3, 2), (2, -1, -3). 4. Where are all the points for which x = 0? y » 0? z = 0? 5. Where are all the points for which x = -2? y - 3? z - -2? 6. From the point (zi, y\, Zi), perpendiculars are drawn to the codrdir nate planes. Find the coordinates of the feet of these perpendiculars. 206. Geometrical methods of finding the coordinates of a point in space. — Since any point P in space c^tn be regarded as the vertex of a rectangular parallelepiped which has the opposite vertex at the origin, the codrdinates of P can be found geometrically in a number of different ways of which the following are the most useful. (1) From P draw a line, Fig. 171, perpendicular to the try-plane and meeting it in K. From K draw a line perpendic- ular to the x-axis and meeting it in H . Then OH - x, HK -•», and KP = z. (2) From P draw planes perpendicular to the x, y, and z-axes, respectively, and let the axes intersect these planes in the points H , M , and L, respectively. Then OH = x, OM = y, and OL = z. (3) From P draw lines perpendicular to the x, y y and z-axes meeting them in the points H } M, and L respectively. Then OH = x, OM = y, and OL = z. 207. Distance between two points. — The distance between two points Pi(xi, j/i, Zi) and P 2 (x 2 , 2/2, Z2) is found by con- structing a rectangular parallelepiped, Fig. 172, having as opposite vertices Pi and P 2 , and whose edges are parallel to the coordinate axes. Then PiP 2 is the length of a diagonal of this rectangular parallelepiped. Digitized by Google 264 ANALYTIC GEOMETRY [§208 Since PiSP* is a right triangle, P l P i t » P^ + SP S *. Since PiRS is a right triangle, P^S* - P\R* + RS*. Therefore iW - PiR* + R& + SPf. Substituting PiP» — d, Pi# = a* — Xi, AS — y* — yi, and SPi = z% — zi y and extracting the square root of both sides of the equation, gives the distance formula [«] d - V(xi - x*) 1 + (yi - y*)* + (ii - z,) 1 . J f p. / s / Pi ^ s^ g ^ r o y /* / / / *x Fio. 172. Fig. 173. 208. CoSrdinates of a point dividing a line segment in the ratio ri to r*. — As in plane analytic geometry, the ratio r P P — = n^ will be considered positive for internal division ft roTt and negative for external division. Let Pi and Pi, Fig. 173, be the end points of the segment and let P be the point of division. Through P it P 2 , and Po draw planes perpendicular to the s-axis meeting it in the points iVi, 2Vj, and iVo respectively. By a familiar theorem in solid geometry, PiPo _ NiNo PoPt ~ NoNi But NiNo = ON - ONi,ON = x©, and ONi = x h Art. 206. Hence iViiVo = xo — Xi. Similarly iVoiV2 = x% — xo. Digitized by Google §208] SOLID ANALYTIC GEOMETRY 265 P P r Substituting these values and replacing p 1 ^ by -\ gives T\ Xo — * Xi Tt X2 — Xo Solving for xo, Xt\ = ■ n + r t By drawing planes perpendicular to the y-axis and the z-axis similar formulas are obtained for y and Zq. Therefore the coordinates of Po dividing the line P\Pi in the ratio r\ to r 2 are taoi ~ riXt + rgXi riy a + r 2 yi w riZ 2 + r^zi 1491 Xo = r l + r 2 ' y ° = r,+r 2 ' Z ° = r> + u ' EXERCISES 1. Find the distance between the points ( — 1, 3, 7) and (1, 9, 16). 2. Find the distance between the points (3, —2, 4) and (6, —6, —8). 3. Show that the points (1, -2, 3), (7, 0, 6), (4, 6, 8) form a right triangle. 4. Show that the points (3, -1, 4), (4, 1, 7), (1, 4, 6) form a right triangle. 5. Show that the points (1, 7, 6), (2, 2, 11), (2, 8, 13) form an isosceles triangle. 6. Show that the points (-4, 2, 5), (-1, 5, 2), (-3, 3, 0) form an isosceles triangle. 7. Show that the points (7, -1, 2), (4, 2, 2), (4, -1, 5), (3, -2, 1) are the vertices of a regular tetrahedron. 8. Find the lengths of the medians of the triangle whose vertices are (-1.7,4)T, (3, -5, -2), (-5, 1,6). 9. Find the codrdinates of the point which divides the line joining (7, 2, 6) to ( -3, 7, -9) in the ratio of 2 :3. 10. Find the codrdinates of the point which divides the line joining (7, -6, 2) to (-3, 4, -5) in the ratio -3 :4. 11. In what ratio is the line joining (2, —6, 3) to (4, —3, —6) divided by the xz-plane? 12. The beginning of the line segment which is divided in the ratio 4:3 by the point (1, 2, -6) is the point (-1, 6, -2). Find the codrdinates of the other extremity. Digitized by Google 266 ANALYTIC GEOMETRY [§209 18. Prove analytically that the straight lines joining the mid-points of the "opposite edges of a tetrahedron pass through a common point and are bisected by it. 14. Prove analytically that the straight lines joining the mid-points of the opposite sides of any space quadrilateral pass through a common point, and are bisected by it. 209. Orthogonal projections of line segments. — In general two lines in space will not intersect.. If parallels to these lines are drawn through any point, the angle made by these inter- secting lines is denned as the angle made by the non-intersecting lines. If through a point P in space a plane is constructed perpendicular to a given line, the point P* where the plane meets the line is defined as the orthogonal projection of the point P on the line. F 174 If the orthogonal projection of the end points Pi and P 2 of a line seg- ment, Fig. 174, on a line I are iY and P 2 ', then the line segment Pi'P 2 ' is said to be the orthogonal projection of the line segment PiP 2 on I. With these definitions it is easy to derive formulas for the projection of a line segment on a given line. Let PiP 2 , Fig. 174, be a line segment of length d, let Pi'P 2 ' be its projection on the line I, and let 6 be the angle between PiP 2 and P/P,'. Through P x draw a line parallel to I and meeting the plane passing through P 2 , perpendicular to line I in the point P 2 ". Join PtPt", P 2 "P 2 ', and PiPi'. Then is the angle P 2 "PiP 2 , and Pi'P 2 ' = PiP 2 " = PiP 2 cos 6 = d cos 0. This gives: Theorem 1. — The projection of a directed line segment on a given line is equal to its length multiplied by the cosine of the angle between the lines. Digitized by Google §210] SOLID ANALYTIC GEOMETRY 267 Another theorem which is useful in solid analytic geometry is the following: Theorem 2. — The projection on any line of *the straight line joining any two points is equal to the algebraic sum of the projection of any broken line z joining these points. Proof.— Let P x P 2t Fig. 175, be the' straight line joining Pi and Pa, let I be the line on which PiP* is to be projected, and let P1P3P4P6P2 be any broken line joining Pi to P*. If the points P/, P 8 ', P/, P 6 ', P* are the pro- jections of the points Pi, P 3 , Pa,y Pby Pi, respectively, then Fio. 175. Proj. P1P3 + proj. PsP* + proj. P4P5 + proj. P^P* = Pi'Pa' + Pz'Pl + Pt'Pt' + Pi'P,' = Pi'P*'. But proj. PiP 2 = PiTY. This proves the theorem. 210. Direction cosines of a line. — Let the angles which any line in space makes with the positive x, y, and 2-axes be re- spectively, a, P, and 7. These angles are called the direction angles of the line. Their cosines, cos a , cos ($, cos 7 are called the " wr direction cosines of the line. If Pi and P 2 , Fig. 176, are any two points on a line and d is the distance P1P2, the direction cosines are given by the formulas [60] cos « = ? 2 -=^» cos = ?*-=^> cos y - -*-^— • Digitized by Google 268 ANALYTIC GEOMETRY [§210 To prove this, let the projection of P\P* on the z-axis be Px'Pi', then Pi'Pt' = dcosa. But Pi'P,' = OP t ' - OPS and OP*' = a* and OPS = x x . Art 206. Therefore Xi — Xi = d cos a, or x 2 — x\ cos a = — 3 a The remaining two formulas are found by projecting PiP* on the y-axis and the z-axis respectively. These three direction cosines are not independent, for squar- ing each equation and adding gives cos 2 a + cos 2 + cos 2 ? = (a* - xQ 2 + (y 2 - y x y + (z t - z x y d* d* d* ' Therefore [61] COS 2 a + COS 2 + COS 2 ? = 1. Example. — Find the direction cosines of a line if they are proportional to the numbers 2, —9, 6. Solution. — Since cob a : cos p : cos y —2: —9:6, cos a = 2k, cos = — 9k, cos y = 6fc. But by [51], the sum of the squares of these cosines equals unity, therefore 4fc» + 81fc» + 36fc* - 1. Or fc= ± IT Substituting, cos a 2 cos/3 = 9 cos y — 6 11 r cos a = 2 cos/3 = 9 = yz > COS y s — 6 11* Digitized by VaOO^ rfe §211] SOLID ANALYTIC GEOMETRY 269 EXERCISES Find the direction cosines of the lines joining the points in exercise 1-3, and the projections of these lines on the three axes. 1. (4, 2, 3) to (5, 3, 4). 3. (-5, 1, 4) to (-3, 4, -2). 2. (-2, 1, 7) to (5, -3, 2). Find the direction cosines which are proportional to the numbers in exercises 4-7. 4. -6, 2, -3. 6. 4, 3, -12. 5. 6, -7, 6. 7. -10, -6, 15. 8. Find the orthogonal projection of the line joining (7, 6, —2) to (5, —3, 4) on the x-axis; on the y-axis; on the z-axis. 9. What are the direction cosines of the x-axis? Of the y-axis? Of the 2-axis? 10. What are the direction cosines of a line parallel to the z-axis? Perpendicular to the a?-axis7 11. Where do all the lines lie for which (a) cos a = J, (b) cos = }, (c) cos a = i and cos = i, (d) cos a — 0, (e) cos a = 1? 12. A line makes an angle of 60° with both the x and the y-axis, what angle does it make with the 2-axis? 13. A line makes an angle of 75° with the s-axis, and 45° with the y-axis, what angle does it make with the z-axis? 14. The equal acute angles which a line makes with the x-axis and the y-axis, are each one-half the angle which it makes with the z-axis. Find the direction cosines of the line. 15. The angles not greater than 90° which ^ a line makes with the x, y, and z-axes are proportional to 1, 2, and 3. Find the direc- tion cosines of the line. \ *s N* 211. Polar coordinates of a point. —If the distance OP, Fig. 177, of a point P from the origin is called p, and if the direction angles of OP are a, 0, and 7, then (p, a, 0, 7) are called the polar coordinates of P. The re- lations between the polar coordinates of P and its rectangular coordinates are obtained by replacing fa, j/i, Zi) of article 210 by (0, 0, 0) and (x 2 , y 2 , z 2 ) by (z, y,z). Fig. 177. Digitized by Google 270 ANALYTIC GEOMETRY Since d « Vz* + V* + «S formula [60] gives the coordinates in terms of the rectangular coordinates. X 11212 polar cosa [52] ±\/x* + y* + z* COS J = 7 y =$ ±Vx* + y* + z* COS y = ±Vx* + y 2 + z 2 Note that the radicals must be taken either all positive or all negative. Replacing y/x 2 + y 2 + z 2 by its value p, and clearing of fractions gives the rectangular coordinates in terms of the polar coordinates. x = q cos a, [53] y - 9 cos ff, z = g cos y. Note that the direction cosines are not independent but are connected by the equation cos 2 a + cos 2 /8 + cos 2 ? = 1. 212. Spherical coordinates. — Another method of locating a point in space is by means of spherical coordinates. From P, Fig. 178, drop a line perpendicular to the xy-plane meeting it in M. Join 0, called the pole, to P, and to M . Then the spherical coordinates of P are p, 0, and <f>, which are written (p, 6, <!>), where p = OP is the distance of P from the origin; 6 = angle NOM is the angle through which the positive x-axis would have to rotate to coincide with OM; and ^ = angle ZOP is the angle which OP makes with the positive &-axis. The quantity p is taken positive if measured along the radius Digitized by Google §213] SOLID ANALYTIC GEOMETRY 271 vector, and negative if measured along the radius vector produced through the origin. The angle 6 can have any value from 0° to 360°. The angle 4> is restricted to values from 0° to 180°. The relations between spherical and rectangular coordinates are then z = q sin <p cos 6, y = p sin y? sin 0, z = q cos <p 9 '= + Vx 2 + y 2 +z 2 , ,-i I [64] tan y - 1 *- = sin~ X ±Vx 2 + y 2 ^= COS" 1 ±Vx 2 + y 2 + z 2 The convention with regard to signs is that either all the upper signs must be used, or else all the lower. If the pole of a spherical coordinate system were taken at the center of the earth, the z-axis passing through the north pole, and the xz-plane passing through the meridian of Greenwich, then the spherical coordinates of a point in the northern hemisphere can be so chosen that p will give the distance of the point from the center of the earth, its longitude and <p its co- latitude. 218. Angle between two lines. — Let the two lines be h and U Fig. 179, with direction angles a h fa, 71, and c*2, &, 72> respectively, and let $ be the angle between h and 1%. In order to find 0, draw two lines OP\ and OP 2 through parallel to h and k respectively, also draw ON, NM, MP h the coordinates of Pi, and let OPi = pi. By article 209, the angle between OP\ and OP 2 equals 6. Fig 179. Digitized by Google 272 ANALYTIC GEOMETRY [8213 Project OPi and the broken line ONMP x on 0P t . By theorem 2, Art 209, proj. OPi = proj. ON + proj. NM + proj. MPi. By theorem 1, Art 209, proj. OP i on OP i = pi cos 0, proj. ON on OP 2 = %i cos as, proj. NM on OP2 = j/i cos ft, proj. JfPi on OP 2 = 21 cos 72. Therefore picos = xicos at + yicos ft + 3icos 72. Replacing x\ f j/i, Si by their equivalents, [53], and dividing both sides of the equation by pi, gives the required expression for 0, [55] cos = cos ai cos a 2 + cos (3i cos § 2 + cos y t cos y 3 . If the two lines are perpendicular to each other, cos ai cos at + cos ft cos ft + cos 71 cos 72 = 0. If the two lines are parallel to each other, it is evident that either a\ = at, ft = ft, and 71 = 72, or ai = 180° — at, ft = 180° - ft, and 71 = 180° - 72. Example 1. — Find the polar coordinates of the point (1, — 1,— V2)- Prom [52], p — y/i — 2, cos a — i, cos — — i, cos 7 = — i\/2 .-.« . 60°, - 120°, 7 - 135°. Then the polar coordinates of (1, -1, — \/2) are (2, 60°, 120°, 135°). If the negative sign is taken with p the polar coordinates are (-2, 120°, 60°, 45°). Example 2. — Find the spherical coordinates of (1, — 1, — V2)- From [64], p = Vi « 2, - tan" 1 (- 1) - sin- 1 3= * 315 °> **<*«> = cos- 1 -^-^? - 135°. Then the spherical coordinates of (1, -1, - y/2) are (2, 315°, 135°). If the negative sign is taken with each of the radicals the spherical coordinates are (-2, 135°, 45°). Example 3. — Find the direction cosines of a line which is perpendicular to two lines having direction cosines proportional to —1, 2, 6 and 1, 4, 3 respectively. Digitized by Google §213] SOLID ANALYTIC GEOMETRY 273 If cos a, cos 0, cos y are the required direction cosines, [55] and [51] give the three equations: —cos a + 2 cos + 6 cos y = 0, cos a + 4 cos + 3 cos y =0, cos* a + cos* + cos* 7 = 1.. Solving these equations, gives cos a'= f, cos = — f, cos y =f, or cos a — — $ , cos = $ , cos y = — f . Example 4. — Find the projection of the line segment U joining the points ( — 1, 3, 6) and (3, 7, — 1) on the line h joining the points (3, 1, —2) and (6, 7, 0). From [50] the direction cosines of h are $, J, — J and of U are f, I ?. If 6 is the angle between the two lines, by [55], 12 + 24 - 14 22 cos $ ,63 63 The length of h by [48] is d = 9, and the projection of U on U by 9 X 22 22 theorem 1, Art 209, is equal to d cos = — ~« — = -=r- EXERCISES Find the polar coordinates of the points in exercises 1-3, if their rectangular coordinates are: I. (1, V2, -1). 2. (4, -4, 4V2). 3. (1, 1, 1). 4. If the polar coordinates of a point are (3, 60°, 60°, y) t find y. Find the spherical coordinates of the points in exercises 5-7, if their rectangular coordinates are: 5. (2, 2V3, 4V3). 6. (-3, -\/3, -2). 7. (-\/6, VS, 2). 8. Find the acute angle between the two lines having direction cosines proportional to 11, —10, 2 and —5, 2, 14. 9. Find the direction cosines of a line which is perpendicular to two lines having direction cosines proportional to 2, 4, —3 and —1, 4, 3, respectively. 10. Find the projection of the line segment joining (3, —1, 4) to (4, 1, 6) on'the line joining (4, 2, -5) to (-2, 4, -2). II. Find the projection of the line segment joining (7, 2, —3) to (2, 4, 3) on the line joining (1, -4, 3) to (7, -11, -3). 12. Find the projection of the line segment joining (2, 1, —3) to (-2, 3, 1) on the line joining (3, -10, 4) to (12, 8, -2). 13. Verify the conventions used with regard to signs in article 212. 18 Digitized by Google 274 ANALYTIC GEOMETRY [§214 SURFACES 214. Locus in space. — If an equation in two variables is given in plane analytic geometry, values can be assigned at pleasure to one of these variables, and then the other is determined. The locus of all points satisfying such an equation is found in general to be a curve. On the other hand, in solid analytic geometry, if an equation in three variables is given, values can be assigned at pleasure to two of the variables and then the third variable is deter- mined. For instance, in the equation z — x 2 + y 2 , to every pair of values of x and y there corresponds a value of z. Hence for every point in the xy-plane there will be a corresponding point in space for the locus of z = x 2 + y 2 . If these points are thought of as a whole, it is obvious that they all lie on a surface. In general then the locus of a single equation in space is a surface. Sometimes one or even two variables may be missing in an equation, in which case such an equation will give rise to a special surface. 215. Equations in one variable. Planes parallel to the axes. — The equation x = a, is satisfied by all values of y and z, since these variables can be regarded as entering into the equation x = a with zero coefficents. Hence all the points satisfying x = a will lie in a plane parallel to the yz-plane and cutting the x-axis at the point x = a. If the equation has the form f(x) = 0, the locus will consist of a series of planes, all parallel to the j/2-plane and cutting the x-axis at points whose abscissas are the roots of f(x) = 0. Like considerations hold for equations which contain only the coordinate y, or only the coordinate z. 216. Equations in two variables. Cylindrical surfaces. — A cylindrical surface is generated by a straight line which moves so as to be always parallel to some fixed line, while intersecting a fixed curve. The fixed curve is called the directrix of the cylindrical surface, and the moving line in Digitized by Google §216] SOLID ANALYTIC GEOMETRY 275 any one of its positions on the surface is called an element of the cylindrical surface. A plane can be regarded as a particular case of a cylindrical surface whose directrix is a straight line. Consider the equation x* + y 2 = 25, Fig. 180. In two dimensional space this is the equation of a circle with center at the origin and radius equal to 5. In three dimensional space, the coordinate z can be regarded as entering the equa- tion with a zero coefficient. Hence with any value of x and y which satisfies the equation, say x = 3 and y = 4, there can be associated any value of z. Thus, the . points (3, 4,-1), (3, 4, 0), (3, 4, 2), and, in general, (3,4,2) where z has any value, will all be points on the surface. These particular points all lie on a line per- FlG# 180, pendicular to the afy-plane and passing through the point 3=3, y=4 in the sy-plane. In like manner through every point on the circle x 2 + y 2 = 25 in the xy-plane there passes a line perpendicular to the xy- plane, and every point of this line satisfies the equation x 2 + y 2 = 25. Hence the locus of the equation x 2 + y 2 = 25 is a cylindrical surface with elements perpendicular to the xy- plane, in other words parallel to the 2-axis, and having the circle x 2 + y 2 = 25 as directrix. Another illustration is the surface z 2 = x. Its elements are parallel to the y-axis and its directrix is the parabola z 2 = x in the a#-plane. This is called a parabolic cylindrical surface. In general an equation f(x, y) = represents in space a cylindrical surface whose elements are parallel to the 2-axis, and whose directrix is the curve f(x, y) = in the sy-plane. The equations f(y, z) = and /(x, z) = represent cylin- drical surfaces similarly situated with reference to the x and the y-axes, respectively. Digitized by Google 276 ANALYTIC GEOMETRY [§217 217. Spheres. — Let C(h, k y I) be the center of a sphere of radius r. Since every point P on the sphere is at the constant distance r from its center CP = r. Or V(x - h) 2 + (y- k) 2 + (z - I) 2 - r. [66] .'. (x - h)« + (y - k)« + (z - 1)« = r*. This is the equation of the sphere with center at C and radius r. 218. Surfaces of revolution. — A surface formed by revolv- ing a curve about a line in its plane is called a surface of revolution. The simplest cases are those where the curve is revolved about one of the coordinate axes. Sup- pose it is desired to revolve the parabola y 2 = x about the x-axis. Let P, Fig. 181, be any point on the parabola. As the curve revolves about the x-axis, P describes a circle of radius NP. When P is in the xy-plane, NP 2 = x. When P takes another position, say P', then NP' = NP and therefore NP' 2 = x, but NP' = \NM 2 + MF 2 = %* + z 2 . Replacing NP' by its value gives (y/y 2 + z 2 ) 2 = x, or y 2 + z 2 = x. Since P' can be any point on the surface, this is the equation 4 of the surface of revolution. This equation was obtained by replacing y by y/y 2 + z 2 . If f(x : y) = is the equation of any curve in the xy-plane which is to be revolved about the x-axis, the same method of reasoning shows that /(x, \/y 2 + z 2 ) = is the equation of the surface of revolution. In like manner the equation of the surface of revolution obtained by revolving /(x, y) = about the y-axis is /(Vx 2 + z 2 , y) = 0. Fiq. 181. Digitized by Google §2181 SOLID ANALYTIC GEOMETRY 277 Thus if y 2 = x is revolved about the y-ax is, the equation of the surface of revolution is y 2 = \/x 2 + z 2 , or y A = x 2 + z 2 . Similar formulas hold true if the curve is given in one of the other coordinate planes and revolved about the corresponding coordinate axes. For instance if the curve /(y, z) = is revolved about the g-ax is, the equation of the surface of revolution is f(\^x 2 + y 2 , z) = 0. If a circle and a line are in the same plane, and the line does not intersect the circle, the surface formed when the circle revolves about the line is called an anchor ring or torus. EXERCISES 1. What is the equation of the plane parallel to the xy-plane and 3 units above it? 4 units below it? 2. What is the equation of the xy-plane? Of the xa-plane? Of the yz-plane? 8. What is the equation of the locus of a point distant 3 units from the x-axis? 4 units from the z-axis? Find the equation of the locus of a point determined by the conditions in exercises 4-9. 4. Equidistant from the points ( — 1, 2, 3) and (3, 4, —2). 5. Equidistant from the xy-plane and the xz-plane. 6. Equidistant from the z-axis and the y-axis. 7. Equidistant from the x-axis and the yz-plane. 8. Equidistant from the point (2, —4, 3) and the x-axis. 9. The sum of the squares of its distances from the point (1, 1, 1) and (2, — 1, 3) is constant and equal to 17. 10. Find the equation of a sphere with center on the x-axis, radius equal to 9, and which passes through the point (2, 4, —8). 11. Find the equation of a sphere with center in the xy-plane, radius equal to 7, and which passes through the points (3, 4, 6) and (7, 3, 3). 12. Find the equation of a sphere passing through the points (2, 3,-6), (5, 3, -5), (5, -2, 10), and (-3, 6, 6). Find the equations of the surfaces of revolution obtained by revolving the curves in exercises 13-24 about the axes as indicated. 18. y = x, about the x-axis. 14. y — s, about the y-suda. 15. y a x 1 , about the x-axis. 16. z* = x, about the x-axis. 17. x* =■ -2«, about the x-axis. Digitized by Google 278 ANALYTIC GEOMETRY [§219 18. x* ■» 2z, about the z-axis. 19. x* — 2« + z % — 0, about the x-axis. 20. a;* — 2s + 2* — 0, about the z-axis. 21. y =■ sin z s about the z-axis. 22. y ■» sin «, about the y-axis. 28. The ellipse - x + ?| » 1, about its major axis. This surface is called a prolate spheroid. 24. The ellipse — t + j- t = 1, about its minor axis. This surface is called an oblate spheroid. 25. What is the equation of the anchor ring obtained by revolving the circle lying in the ay-plane, with center at the point (0, 4) and radius 2, about the x-axis. Skeu;h and describe the following surfaces. 26. x - y = 0. 81. a* + 4z l - 4. 27. x* + y* - 4. 82. a* - 4z» - 4. 28. x* - 2x + y« - 0. 88. a* + y* + z* - 2z - 2y - 2* - 6 = 0. 29. y 2 - 2y + *« = 0. 34. a* - 3x + 2 - 0. 80. y* - 2« - 0. 85. y* - 1 = 0. CURVES IN SPACE 219. Equations of curves. — Since a single equation in three dimensional space is the equation of a surface, two equations will be satisfied simultaneously by all the points lying on the intersections of the two surfaces. In other words, it takes two equations in solid analytics to define a curve. Thus, y = is not sufficient to define the equation of the x-axis, for every point in the zz-plane satisfies this equation. Neither is z = sufficient, for this is satisfied by every point in the xy-plane, but y — and 2 = are satisfied only by those points common to the sz-plane and the sy-plane, namely, the z-axis. Therefore, y = and z = are the equations of the x-axis. Similarly x = and y = are the equations of the 2-axis; and x = and z = are the equations of the y-axis. Since it is evident geometrically that an unlimited number of surfaces can be passed through any curve, and that any two of these surfaces will be sufficient to define the curve, any Digitized by Google §220] SOLID ANALYTIC GEOMETRY 279 space curve can be represented by an unlimited number of pairs of equations. Thus a circle in space has for its pair of equations, the equations of any two spheres passing through it, or the equation of any sphere and the equation of the plane in which the circle lies. Even this does not exhaust all the possibilities of representing a circle, since any two surfaces passing through the circle will define it. 220. Sections of a surface by planes parallel to the coSrdi- nate planes. — The curve in which a surface is cut by a coordi- nate plane is called the trace of the surface in the coordinate plane. Thus, the sphere x 2 + y 2 + z 2 = 25 and the plane 2 = define a curve, the circle formed by the intersection of the sphere and the xy-plsme. If z is put equal to zero in the equation of the sphere, it becomes x 2 + y 2 = 25. This is the equation of the trace of the sphere in the xy-plane. By putting y = or x = the trace of the sphere in the sz-plane or the $/3-plane is obtained. Consider the two equations x 2 + y 2 + z 2 = 25 and z = 3. The curve AB, Fig. 182, com- ., % mon to these two surfaces is known from solid geometry to be a circle. If 3 = 3 is substi- tuted in x 2 + y 2 + z 2 = 25, it becomes x 2 + y 2 = 16. This is the equation of a circular cylinder. 1 Since every point satisfying the equation of the sphere and the plane satisfies the equation of the cylinder, this cylinder must pass through the circle AB. Hence substituting z = 3 in the equation of the sphere x 2 + y 2 + z 2 = 25, gives the equation of a cjdinder passing through the intersection of the plane 2 = 3 and the sphere x 2 + y 2 + z 2 = 25. 1 For brevity the words cylindrical surface are often replaced by the word cylinder. Digitized by Google 280 ANALYTIC GEOMETRY [§221 The meaning of this substitution can be regarded from another standpoint. The equation x 2 + y 2 = 16 is the equation of the circle CD in which the cylinder cuts the ay-plane. All the elements of the cylinder are perpendicular to the £t/-plane and pass through the circle AB. Hence the circle CD is the projection of the circle AB in the xy-plane. In other words, substituting z = 3 in the equation of the sphere x 2 + y 2 + z 2 = 25, gives the equation of the projection on the xy-plane of the curve common to the plane z = 3 and the sphere x 2 + y 2 + z 2 = 25. In general, the substitution z = c, where c is some constant, in the equation of a surface can be regarded either as giving the equation of a cylinder passing through the intersection of z = c and the surface or as giving the equation of the projection on the #t/-plane, of the curve of intersection of the plane z = c and the surface. By giving c different values, the shape of different cross sections of the surface in planes parallel to the xy-plane are obtained. Like considerations hold for the substitution of x = a or y = b in the equation of a surface. 221. Projections of curves on the coordinate planes. — When a curve is defined in space by two equations, it is desirable sometimes to know what are the equations of its projections in the three coordinate planes. Consider the curve defined by the equations x 2 + y 2 + z 2 = 49, (1) x 2 + 3y 2 - z 2 = 39. (2) If z is eliminated between these two equations, the resulting equation 2x 2 + 4y 2 = 88, or x 2 + 2y 2 = 44 (3) represents an elliptical cylinder. Furthermore any point whose coordinates satisfy equation (1) and (2) will also Digitized by Google §221] SOLID ANALYTIC GEOMETRY 281 satisfy equation (3), therefore the cylinder (3) passes through the curve defined by the equations (1) and (2). From another standpoint x 2 + 2y 2 = 44 is the equation of the directrix of the cylinder x 2 + 2y* = 44, and since the elements of this cylinder are perpendicular to the sy-plane, the equation x 2 + 2y 2 = 44 is the equation of the projection on the sy-plane of the curve defined by equations (1) and (2). In general, to find the equation of the projection on the xy- plane of the curve defined by the equation f\(x, y,z) = and f2(x y y, z) = 0, elminate z between these two equations. The resulting equation g(x, y) = is the equation of the projection on the xy-plane of the curve that is defined by the equations /i(z, y, z) = and f 2 (x, y, z) = 0. Proof. — Every point which satisfies simultaneously the equations fi(x } y } z) = and / 2 (x, y, z) = will also satisfy g(x, y)=0 and therefore g(x } y) = will pass through the inter- sections of these two surfaces. But g(x } y) = is the equation of a cylinder whose elements are perpendicular to the xy-plane. At the same time g(x f y) = is the equation of the trace of this cylinder in the ;ry-plane. Therefore, g(x y y) = is the equation of the projection on the xy-plane of the curve defined by the equations fi(x, y, z) = and/ 2 (x, y, z) = 0. In like manner it can be shown that to find the equation of the projection on the xz-plane of a curve defined by the equations fi(x, y, z) = and / 2 (x, y, z) = 0, eliminate y between these two equations; and to find the projection on the yz-plane eliminate x. For example, the projection on the x?/-plane of the curve defined by equations (l) and (2) is the ellipse x 2 + 2z % = 54, and projection on the 2/z-plane is the equilateral hyperbola z % — y* = 5. EXERCISES Discuss and draw the traces on the three coordinate planes of the surfaces in exercises 1-3. 1. x* + 2y* + 3s* = 6. 2.^+^+2-0. S. x* + y* - z - 1. Digitized by Google 282 ANALYTIC GEOMETRY [§222 Find the equations of the projections on each of the three coordinate planes of the curves in problems 4-9. 4. x* - y* + z* - 4, 7. x* + y* - a*, 2x» + y* - &* - 6. , s» + 2» - a*. 5. x* - y - 0, 8. x* + y» - a*, 2x* + y — * f ■■ 0. 2 = mx. 6. x* + y* + 2* - a*, 9. y* + z* - 4as, ** + y 1 — ox. y* = ox. 10. Show that sections of - - t — 1^- = z are hyperbolas if perpendicular to the z-axis, but parabolas if perpendicular to the x-axis or y-axis. DISCUSSION OF EQUATIONS OF SURFACES 222. Surfaces in space. — It is much more difficult to vis- ualize a surface in solid analytic geometry from its equation than to visualize a curve in plane analytic geometry from its equation. The following discussion similar to the one in the plane case is helpful. (1) Symmetry. (2) Intercepts on the axes. (3) Traces on the coordinate planes. (4) Sections of the surface by planes parallel to the coordi- naie planes. See Art. 220. (1) Symmetry. — To test the symmetry of a surface with respect to the coordinate planes, (a) replace x by —x, (6) replace y by -y, (c) replace z by —z. If the equation of the surface remains unchanged in case (a) it is symmetrical with respect to the j/z-plane, in case (6) with respect to the £2-plane, in case (c) with respect to the zy-plane. To test for symmetry with respect to the axes (a) replace y by — y and z by —z } (6) replace z by — z and x by — x, (c) replace x by — x and y by —y. Digitized by Google §222] SOLID ANALYTIC GEOMETRY 283 If its equation remains unchanged in case (a) it is symmetri- cal with respect to the x-axis, in case (6) with respect to the t/-axis, in case (c) with respect to the z-axis. To test for symmetry with respect to the origin replace s by -x, y by -j/, z by -z. If its equation remains unchanged, ita surface is symmetrical with respect to the origin. (2) Intercepts on the axes. — To get the intercepts on the x-axis, set both y and z equal to zero in the equation of the surface and solve the resulting equation for x. The solutions of this equation are the intercepts on the x-axis. Similar considerations hold true for the t/-axis and the 3-axis. (3) Traces on the coordinate planes. — To get the trace of a surface in the xy-plane set z = 0. The resulting equation is the equation of the trace of the surface in the xy-plane. Similar considerations hold true for the traces in the sz-plane and the ys-plane. See Art. 220. Example. — Discuss and draw the locus of the equation x 2 + 2y 2 = z. (1) This surface is symmetrical to the ys-plane, the zz-plane and the s-axis. (2) Its intercepts on the three axes are 0. (3) Its traces are as follows: In the xy-plane, the point ellipse x 2 + 2y 2 = 0. In the xz-plane, the parabola x 2 — z. In the yz-plane, the parabola 2y 2 = z. (4) Taking sections by planes z = c shows the projections of these sections to be the ellipses ?!.!£._ -I Fig. 183. c + ic ~ 1 ' If c<0, these ellipses are all imaginary, hence no part of the surface lies below the xy-plane. If c — 0, the equation x 2 + 2y 2 = 0, shows the section to be a point ellipse. As c increases from without limit, the ellipses increase in size without limit, the semimajor axis being \fc and the semiminor axis i\/2c, hence the surface is as pictured in Fig. 183. In this case it is not necessary to take sections parallel to the other coordinate planes. Digitized by Google 284 ANALYTIC GEOMETRY [§223 QUADRIC SURFACES OR CONICOIDS 223. General equation of second degree. — The locus of the general equation of the second degree, Ax 2 + By 1 + Cz* + Dxy + Eyz +Fxz+Gx+ Hy + Kz + L = 0, is called a quadric surface. It is also called a conicoid because every section of a quadric surface by a plane is a conic. By rotation and translation of axes it can be shown that this equation has for its real locus, five distinct types of surfaces besides cylinders, cones and degenerate forms like planes, lines and points. These five types will now be considered. 224. Ellipsoid, g + g + g-i. (1) This surface, Fig. 184, is symmetrical to all the coordinate planes, all the coordinate axes, and the origin. (2) Its intercepts on the axes are x = ±a, y = ±b, z = ±c. (3) Its traces are as follows: i Z \^ X Y* \- r — J Fiq. 184. = 1. In the zy-plane, the ellipse In the xz-plane, the ellipse In the 2/z-plane, the ellipse j- 2 + - 2 = 1. o c (4) Sections of the ellipsoid by the planes z = k are the ellipses a 2"1- fe 2 l C 2, Z -*> X a a or + b\ = 1, z = k. . (C 2 _• fc 2) - (C 2 _ J,) Digitized by Google §225] SOLID ANALYTIC GEOMETRY 285 These ellipses have their centers on the s-axis, semimajor axes equal to -\c 2 — fc 2 , and semiminor axes equal to - Vc 2 — k 2 . Here it is assumed that a > b. If a < 6, the axes c are interchanged. As k increases numerically from to c, the axes decrease from a to 0, and from b to 0, respectively. When k is numeri- cally greater than c, the ellipses become imaginary. Hence the ellipsoid is contained between the planes z = — c and z = c. A similar discussion for the other axes shows that sections parallel to the other coordinate planes are ellipses, and that the ellipsoid is contained between the planes y = — 6 and y = b, and between the planes x = — a and x = a. The surface can be thought of as generated by a variable ellipse moving parallel to the zy-plane, with its center always on the s-axis, and the end points of its axes always on the ellipses x 1 z 2 , y 2 z 2 -, + -,= l,and^ + - 2 = l. Special forms of the ellipsoid are the prolate spheroid when b = c and a>b, and the oblate spheroid when b = c and a<b. x 2 y 2 z 2 225. The hyperboloid of one sheet — 2 + ^- 2 5 = !• (1) This surface, Fig. 185, is symmetrical to all the coordi- nate planes, all the coordinate axes and the origin, (2) Its intercepts on the axes are x = ±a, y = ±6. (3) Its traces are as follows: X 2 y 2 In the a?2/-plane, the ellipse "2 + fr = 1 • x 2 z 2 In the xz-plane, the hyperbola — 2 2 = 1. *2/ 2 z 2 In the 2/s-plane, the hyperbola r^ 2 = 1. fe 2 Digitized by Google 286 ANALYTIC GEOMETRY [§225 (4) Sections of the surface by the planes z = k are the ellipses x 2 * 2 + V* - i + fc2 or + y 2 r,{c 2 + fc 2 ) b 2 - 2 (c* + k 2 ) z •■*= k. = 1, z = k, These ellipses are real for all values of fc, increasing in magnitude as k increases numerically from to «. The smallest ellipse is the one for which k = 0, and this is the trace in the xy-plane. The intersections in planes parallel to the other axes are hyperbolas. This surface can be thought of as generated by a variable ellipse moving parallel to the xt/-plane, with its center on the z-axis, and the end points of its axes on the hyperbolas x 2 z 2 A y 2 z 2 - 2 -- 2 =l,andp--- 2 = 1. The hyperboloid of one sheet Fig. 185. has the property that through every point on its surface there can be drawn two lines which lie wholly in the surface. The surface can be covered with a net work of two sets of lines. No two lines of the same set intersect each other, but any line of either set intersects every line of the other set. This surface can be generated by a line moving in such a way that it always intersects three other non-intersecting lines in space. It is called a ruled surface, because through every point on its surface, there can be drawn at least one line which lies wholly on the surface. Digitized by Google §226] SOLID ANALYTIC GEOMETRY 287 226. The hyperboloid of two sheets. - *- - - = 1. »2 K2 r 2 ** a* b 2 c* (1) This surface, Fig. 186, is symmetrical to all the coordi- nate planes, all the coordinate axes, and the origin. (2) Its intercepts on the a>axis are ±a. The intercepts on the y-axis and the z-axis are imaginary. (3) Its traces are as follows: /»»2 A*2 In the a^-plane, the hyperbola — 2 — ~ = 1. In the zs-plane, the hyperbola — 2 ^ = 1. In the ys-plane, the imaginary ellipse ^ + -5 = — 1. c ♦-X Fig. 186. (4) Although the trace in the ys-plane is imaginary the form of the equation suggests that sections parallel to the j/2-plane might be ellipses. Since it is easy to picture a surface in terms of increasing or decreasing ellipses, sections will be taken parallel to the #z-plane. Sections of this surface by such planes parallel to the j/z-plane as x = k, are the ellipses or 2/ a S< fc2 -« 2 > t + z l = *_ 2 - 1 ft* T C 2 a 2 l > - + *-* X = fc, # 2 - ^) -= 1, x = fc. Digitized by Google 288 ANALYTIC GEOMETRY [§227 These ellipses are imaginary if —a<k<a. Hence there is no surface between the planes x = — a and x = a. As k increases numerically from a to », the ellipses increase indefinitely in magnitude. Sections by planes parallel to the other axes are hyperbolas. The surface can be thought of as generated by a variable ellipse moving parallel to the ys-plane, with its center always on the x-axis, and the end points of its axes on the hyperbolas x 1 y 2 , x 2 z 2 7#- Si- ^ and ^-ri= 1. a a' 227. Elliptic paraboloid. ^ + j£ =z. (1) This surface, Fig. 187, is symmetrical to the j/s-plane, the sz-plane, and the s-axis. (2) Its intercepts are x = 0, y = 0, and s = 0. (3) Its traces are as follows: X 2 y 2 In the sy-plane, the point ellipse —^ + p = 0. In the zz-plane, the parabola x 2 = a 2 z. In the ys-plane, the parabola y 2 = b 2 z. (4) The trace in the sy-plane sug- gests that sections parallel to this plane might be ellipses, in fact, the sections of this surface by the planes, z = k are the ellipses ~2 + P " fc > * - *■ +-X or x 2 y 2 a 2 k ^ b 2 k = 1, z = fc. Fig. 187. If fc < 0, the ellipses have an im- aginary locus, hence no part of the surface lies below the xy-plane. As k increases from to a>, the ellipses increase in size indefinitely. The sur- face can be thought of as generated by a variable ellipse igitized by Google §228] SOLID ANALYTIC GEOMETRY 289 moving parallel to the xy-plane whose center is on the 2-axis and the end points of whose major and minor axes are on the parabolas x 2 = a 2 z and y 2 — b 2 z. x 2 y 2 228. Hyperbolic paraboloid. -, — ^ = z. (1) This surface, Fig. 188, is symmetrical to the yz-plane, the 33-plane, and the z-axis. (2) Its intercepts are x = 0, y = 0, and z = 0. (3) Its traces are as follows: x 2 y 2 In the sy-plane, the two lines -| — j^ = 0. In the sz-plane, the parabola x 2 = a 2 «. In the yz-plane, the parabola y 2 = — b 2 z. Fig. 188. (4) Since no trace suggests an ellipse, and since it is easier to think in terms of moving parabolas instead of moving hyper- bolas, sections are taken by planes parallel to the yz-plane. Sections of the surface by the planes x = k are the parabolas These are parabolas, symmetrical to the xz-plane, opening (k 2 \ k, 0, -i) lying on the trace x 2 = a 2 z of the hyperbohc paraboloid in the xz-plane. All of these parabolas are congruent. Hence the hyperbolic 10 Digitized by Google 290 ANALYTIC GEOMETRY [§229 paraboloid may be thought of as generated by a parabola opening downward of latus rectum 6 2 , moving with its vertex on x 2 = a 2 z so that its plane is always parallel to the t/z-plane. Sections parallel to the xz-plane are parabolas opening upward, and sections parallel to the rcy-plane are hyperbolas. This hyperbolic paraboloid has the property that through every point on its surface there can be drawn two lines which lie wholly in the surface. The surface can be covered with a network of two sets of lines. No two lines of the same set intersect each other, but any line of either set meets every z line of the other set. Hence the hyperbolic paraboloid is also a ruled surface. This surface can be generated by a line moving always parallel to a fixed plane, while always intersecting two "*" x non-intersecting lines in space. 229. Cone. a 5 + U2 ~2 ~~ 0* (1) This surface, Fig. 189, is sym- metrical to the three coordinate planes, the three coordinate axes, Fig. 189. an( j ^he origin. (2) Its intercepts on the axes are x = 0, y = 0, and 2 = 0. (3) Its traces are as follows: x 2 v 2 In the xy-plane, the point ellipse -, + ^ = 0. x 2 z 2 In the xz-plane, the two lines —„ . = 0. * a 2 c 2 v 2 z 2 In the yz-plane, the two lines ^ 2 = 0. (4) Sections of the cone by the planes z = k are the ellipses x 2 y 2 a 2 k 2 "*" b 2 k 2 = 1. Digitized by Google §229] SOLID ANALYTIC GEOMETRY 291 As the numerical value of k increases from to », the ellipses increase in magnitude indefinitely. Hence the surface can be thought of as generated by an ellipse moving parallel to x* z 2 the xy-plane, with the ends of its axes in the lines -j * = 0, CL C XI 2 Z 2 and the lines p , = 0. This cone is also a ruled surface, but it is covered by a single set only of lines, all of which pass through the origin. EXERCISES Discuss and draw the surfaces in exercises 1-15. 8. x» + y* « 4«. 9. x» - y* = 4z. 10. x* + y* - z* « 0. 11. xy H- xz H- yz — 0. 12. cos a = 0. 13. cos = £• 14. cos = £• T *?"?~S" a 16. cos* = f Discuss and draw the curves or straight lines in exercises 16-20. 16. x = 3, y = -2. 19. x = y = 2. 17. cos a = cos = 0. 20. cos = £, cos </> *» J. 18. y - x, x 2 + y* - 4. 21. Find the equation of the locus of a point which moves so that the sum of the squares of its distances from the x and the y-axis equals 4. Discuss and draw the locus. 22. A point moves so that the sum of the squares of its distances from two fixed points is constant. Prove the locus to be an ellipsoid. Suggestion. — Take the line through the two points to be the x-axis, and a point midway between them as the origin. 23. A point moves so that the difference of its distances from two fixed points is constant. Prove the locus to be an hyperboloid. 24. Find the locus of a point equidistant from the point (p, 0, 0) and the xz-piane. 1. 2. x» + 4*» + 4z» — 9. 9. 3. X* -4y» - 4z* — 9. 4. X* 9 !/* z 2 ^ 4 ^16 1. 5. X 2 9 +*- z* "16 1. 6. X* 9 4 *" "16 1. Digitized by Google 292 ANALYTIC GEOMETRY [§230 THE PLANE IN SPACE 230. Equation of a plane. — Some of the conditions that determine a plane in solid geometry are three points in the plane, or a point and a line in the plane. Unlike the straight line in two dimensional space, these simple conditions do not lend themselves readily to deriving the equation of a plane in three dimensional space. * Rather, one of the simplest ways of deriving the equation of a plane is by using the length of the perpendicular from the origin to the plane and the direc- tion cosines of this perpendicular. This perpendicular is called the normal to the plane. 231. General equation of a plane. — Every equation of the first degree in x, y, and z as [57] Ax + By + Cz + D = represents a plane. Let P\{x\j yi, Zi) and P 2 (x 2 , y 2 , z 2 ) be any two points whose coordinates satisfy [57]. Then Axx + By x + Cz x + D = 0, (1) and Ax 2 + By 2 + Cz 2 + D = 0. (2) Take any two constants r x and r 2 , multiply equation (1) by rx + r 2 multiply equation (2) by rx + r% , and add, A rxx 2 + r t xx , fi r x y% + r 2 y x c r x z 2 + r&x , D = Q rx + r 2 rx + r 2 rx + r 2 This shows that any point on the line joining PxP 2 also satisfies equation [57]. Since Pi and P 2 are any two points on the surface [57], this shows that every line joining two * The equation of a plane through three points can be expressed in determinant form. If the three points are Pi, P s , and Ps, the equation is x y z 1 xi yi zi 1 m x% y% zt 1 x% y% z% 1 Digitized by Google §232] SOLID ANALYTIC GEOMETRY 293 points on the surface lies wholly in the surface, and since this property is characteristic of the plane alone Ax + By + Cz + D = is the equation of a plane. 232. Normal form of the equation of a plane. — Let the length of the perpendicular OR, Fig. 190, from the origin to the plane be p, and let its direction angles be a, 0, 7. If P is any point in the plane, the projection of OP on OR will be constant and equal to p. By theorem 2, Art. 209, the projection of OP equals the sum of the projections of the broken line ONMP on OR. Therefore proj. ON + proj. NM + proj. MP =* p. But proj. ON on OR = x cos a, proj. NM on OR = y cos 0, proj. MP on OR = z cos 7. Substituting these gives [68] x cos a + y cos ff + z cos y = p. This is called the normal form of the equation of a plane. In article 231, it was shown that every equation of the first degree in x, y and z is the equation of a plane. This article proves the converse of that theorem, namely, that every equation of a plane is of the first degree in x> y y and z. 233. Reduction of the equation of a plane to the normal form. — The equations Ax + By + Cz + D = and x cos a + y cos ft + z cos 7 — p = Fig. 190. Digitized by Google 294 ANALYTIC GEOMETRY [§234 will be the equations of the same plane if they differ only by a constant factor. Suppose that k is such a factor, then kAx + kBy + kCz + kD - 0. (1) And therefore kA = cos a, kB = cos j8, kC = cos 7. Squaring each equation and adding gives k 2 (A 2 + B 2 + C 2 ) = 1. <* * = ± y / A 2 +B * + & Substituting this value of k in equation (1), gives Ax + By + Cz + D ±Va* + b* + c* where COS a = ±Va 2 + b 2 + c 2 [59] o B 1 J cos 3 = , ^ > ±Va 2 + b 2 +c 2 c cos y ±VA 2 + B 2 +C 2 -D # p ±Va 2 + b 2 + c 2# 234. Intercept form of the equation of a plane.— -Let the plane cut the s-axis in the point where x = a, the y-axis in the point where y = 6, and the z-axis in the point where z = c. These three quantities are called the intercepts of the plane on the axes. If they are given and none of them is zero, the plane is uniquely determined, for this is equivalent to giving the three points on the plane (a, 0, 0), (0, 6, 0), (0, 0, c). To find the equation of the plane, substitute these three coordi- nates in succession in the general equation Ax + By + Cz + D = 0. (1) Digitized 'by Google §235] SOLID ANALYTIC GEOMETRY 295 This gives the three equations: Aa + D = 0, Bb + D = 0, Cc + D = 0. From which a o c Substituting these values in (1), a o c Dividing by — D, °> i+i+S-«- (2) This is called the intercept form of the equation of a plane. Note that this form is not valid if any of the intercepts are 0, that is if the plane passes through the origin. 235. Equation of a plane determined by three conditions. The general equation of a plane, Ax + By + Cz + D = 0, involves four constants, A, B, C, and D. Any three con- ditions that determine a plane give three relations between these four constants. These three equations can be solved for three of the constants in terms of the fourth providing that the fourth is not zero. Then after substitution the equa- tions can be divided through by the fourth constant as in equation (2), Art. 234. If the fourth constant should be zero, the three equations will turn out to be inconsistent. In such a case solve the equations in terms of another constant. 236. Angle between two planes. — Let the two planes be Aix + Biy + dz + Di = 0, and A t x + B 2 y + C 2 z + D t = 0. Digitized by Google 296 ANALYTIC GEOMETRY [J237 The angle 6 between these two planes is the angle between their normals. Hence by [66] and [69], [61] cos 6 = AxA, + BA + C1C ± VAx* + Bx* + d* V A,* + B,« + <V The two planes are perpendicular to each other if AiA t + BiBi + CiC, - 0. The two planes are parallel if their* normals have the same direction cosines, that is. if ~r = ^ = tt' A% D% Kj% 237. Distance from a point to a plane. — Let Pi(xi, yi, z x ) be the given point and Ax + By + Cz + D = be the given plane. Pass a plane through Pi parallel to the given plane, and find the difference between the normals to the planes. It is then found that the distance d is given by the formula [62] d .A»i + 3ri + tei + D ±Va* + b 2 + c* where the sign is chosen to make 4 positive. Example 1. — Find the equation of a plane passing through the points (2, 1, 7) and (4, —1,-2) at a distance 2 from the origin. Use the normal equation of a plane, x cos a+ y cos 4- * cos y — p « 0, Since the distance of the plane from the origin equals 2, p=2. Since the plane passes through the points (2! 1, 7) and (4, —1, —2). 2 cos a + cos + 7 cos y — 2 »» 0, 4 cos a — cos — 2 cos y — 2 «■ 0. Solving these equations with the identity cos* a 4- cos* 4- cos* 7 — 1, gives cos a = $, cos = — f, cos 7 « f, or cos a = H , cos = Hi cos 7 - -rh- Therefore, there are two solutions to this problem and they are Zx - 6y 4- 2z - 14 - 0, and 105s 4- H4y - 2z - 310 = 0. Digitized by Google J237J SOLID ANALYTIC GEOMETRY 297 Example 2.— Find the equation of the plane bisecting the angle between the planes 3s — y + 2z — 4, and 2x + dy — z — 4. If P(x, y) is any point on the bisecting plane, its distance from each of the two planes is the same. Equating these distances gives 3s - y + 2z - 4 2x + 3y - z - 4 ±Vl4 " ±Vl4 This gives the two planes x - 4y + 3* -0, and 5s+2y +2-8-0. EXERCISES Write the equations of the planes in exercises 1-4 in the normal form and the intercept form. / 1. x - 2y - 2« - 4. 8. 4s + 7y - 4z + 3 - 0. 2. 2x + V - 2z - 9. 4. 12s - y + 12s - 18. Find the equations of the planes which satisfy the conditions of the exercises 5 to 16. 5. Passing through the points (1, 1, 1), (-3, 3, 8), (-2, -3,-2). 6. Passing through the points (2, -1, 0), (4, —2, 4), ( — 1, 3, -1). 7. p = 5, cos a - J, cos = — }. 8. a =» J, b = -}, c - 2. 9. Passing through the points (4, 0, —1), (6, 3, 3) at a distance 2 from the origin. 10. Passing through the point (1, —2, 1) and parallel to the plane y- Sx +4« - 5 = 0. 11. Passing through the points (1, 1, 1), (2, — 1, 2) and perpendicular to the plane 3x + 4y - 7z + 10 = 0. 12. Passing through the point (—2, —1, 3) and perpendicular to each of the planes 2x — 2y — 7z + 3 =» 0, and 4x + y — 4z — 1=0. 13. Passing through the point (1, 1, 2) and perpendicular to the line joining (3, -4, 2) to (4, -6, 3). 14. Perpendicular to the line joining (7, —6, 3) to (1, 2, —5) at its middle point. 15. Parallel to the x-axis and passing through the points (2, 1, 2) and (-3,5,5). 16. Having the foot of the normal from the origin at the point (-3, 4. -2). 17. Find the distance from the point (3, —4, 2) to the plane 5x - 2y - 142 + 15 - 0. Digitized by Google 298 ANALYTIC GEOMETRY [|238 Find the angles between the planes in exercises 18-20. IB. x — y+s«7 and x + y + hz « 3. 19. 2x + y - * « 5 and 4x — 2y — 2z « 3. *>. x + 2y — z - 7 and 2x — y + 7* « 10. Find the equations of the planes bisecting the angle between the planes in exercises 21-23. 21. 2x + V — 2z«l and 3x + 6y — 2s « 7. 22.x+V+« — 4 and 6x — y — z — 2. 23. 2x - y - * « 3 and 5x — 5y + 2z * 4. 24. Determine k so that fcr+0y — 7z— 22=0 shall be two units from the origin. 25. Find the point of intersection of the planes 3x + V - * - 3, x + 5* + 7* - 11, 4z + lOy - 3* - -8. 26. At what acute angle does the plane 2x + Sy + 6* = 3 cut each of the coordinate planes? 27. At what acute angle does the plane 2x + 3y + 6* - 3 cut each coordinate axis? 28. Prove that the planes 2x - y + 3* = 4, a? + 6y - 6z - 5, Sx + 9y - 3« - 22, have a line in common. THE LINE IN SPACE 238. Two plane equation of a straight line. — In article 219, it was seen that it takes two equations in three dimen- sional space to define a curve. Hence the two equations Aix + B lV + dz + Z>x - 0, (1) A& + B 2 y + C* + D* = 0, (2) are the equations of a straight line. Equations (1) and (2) are the equations of any two planes through the line. 239. Projection form of the equation of a straight line. — By eliminating in turn z, y, and x between equations (1) and (2), Art. 238, the equations of the projections of the straight Digitized by Google §240] SOLID ANALYTIC GEOMETRY 299 line on the xy, zz, and yz-planes are obtained. If these equations are l\x + hy +h =0, 7H\X + wi& + Ms = 0, niy + n 2 z + n 8 =0, any two of these equations are the equations of the straight line, and any two of these equations are called the projection form of the equations of a straight line. 240. Point direction form of the equation of a straight line. Symmetric form. Case I. — The line is not parallel to any coordinate plane. Let Pi(xi, 2/1, Zi) be the point and let the direction of the straight line be given by its direction cosines, cos a, cos and cos 7. Then, if P(x, y, z) is any point on the straight line, and d is the distance from Pi to P, by [50] x — xi Q y — y\ z — zi , 1N COS a = -j — y COS fi = - - y COS 7 = -5 (1) Solving each equation for d and equating the results, [63,] *^* = *^£ = x -=*. 1 cos a cos g cos 7 If cos a, cos j8 and cos 7 are replaced by any quantities I, m, n proportional to them, the equation can be written [63d ^ = ^ = ^ 1 m n Case II. — The line is parallel to one or two coordinate planes. Suppose the line is parallel to one of the coordinate planes, say the 2/2-plane, but is not parallel to one of the coordinate axes, then cos a = 0, cos 5* 0, cos 7 -^ 0. Equations [63J and [63 2 ] are not valid, but equation (1) can be written = *Jjp, cos0 = V -=^, cost = *-=p,. giving the equations of the line to be x — Xi = 0, y - y x z - z x and cos jS cos 7 /Google Digitized by* 300 ANALYTIC GEOMETRY [§241 If the line is parallel to two coordinate planes, it is parallel to one of the coordinate axes. If this is the z-axis, then cos a = cos fi = 0, cos 7 = 1. Equation (1) can then be written - £^* = y -^-S cos 7 = *^ a a a giving the equations of the line to be x — Xi = 0, y - y\ = 0. Like considerations hold if the line is parallel to any of the other coordinate axes or planes. 241. Two point form of the equation of a straight line. Case I. — The straight line is not parallel to any coordinate plane. Let the two points through which the line passes be Pi(xi, y x , Zi) and P%(xt f y if s 2 ). Since the direction cosines of this line are proportional to x% — Xi, y% — yi, and z% — Zi, the quantities I, m, and n of [63 2 ] can be so chosen that I = x t - xi, m = j/2 - yi, n = z 2 - z if which gives f64i x ~ Xi _ y-yi . «-«i xj-xi yt-yi zj- zr Case II. — The line is parallel to one or two codrdinate planes. The discussion is similar to that given in article 240. Example 1. — Reduce the equations that define the straight line, 3s + 2y - 2z + 2 = 0, and 6x + 7y - 6s - 3 - to the symmetric form. Solution. — Reduce these equations to the projection form by first eliminating x and then z giving 3y - 2z - 7 - 0, 3* - y + 9 - 0. Solving each for y and equating, 2z + 7 3s + 9 = y - =-~ This can be written in the form x -1-3 2+1 = y » * /Google Digitized by* §241] SOLID ANALYTIC GEOMETRY 301 In order that th e denomina tors shall be direction cosines, multiply each equation by Vi 4- 1 + i " V* an< ^ tne equation becomes s + 3 j/_ * + j A "A" A ' This shows that the line passes through the point (—3, 0, — I) with direction cosines A, -ft, A-. If desired, the point (—3, 0, — 1) can be replaced by any other point on the line, say (—2, 3, 1), in which case the equation of the line takes the form a? + 2 y - 3 z - 1 A " A " A ' Example 2. — Find the equation of a plane passing through the line 2s -3 y-6 2 + 2 . — j — — s —= — = — g— f and the point ( — 1, — 1, —6). Solution. — This equation is equivalent to the two equations 2x - 3 y-6 jtf-6 2 + 2 —4 5- ,and -5^ 3— Simplifying lOx - 4y + 9 - 0, and Zy - 5z - 28 = 0. These are the equations of two planes passing through the given line. The equation of any plane through the line of intersection of these two planes, and hence through the given line, is evidently 10s - 4y + 9 + HZy - 52 - 28) - 0. To make this plane pass through the point ( — 1, — 1, —6), substitute these coordinates and solve for k. The result is k = 3. Hence the required plane is 10* - 4y + 9 + 3(3y - hz - 28) - 0, or 2x + y - Sz - 15 - 0. EXERCISES Find where the lines in exercises 1-5 intersect the three coordinate planes. 1. 4x + j/+2-5=0, 2s-y + 2-l-0. 2. x - y + z - 5, 5x - 6y + 4« = 28. 3. 4x + j/ - 62 =10, 7* + 3y - 82 - 15 - 0. 4. 4s + 3y + 22 - 2, -3* + 4y + 2 - 6 « 0. g -4 _ y + 1 _ 2-3 •• 2 ~ -1 "" 2 ' Digitized by Google 302 ANALYTIC GEOMETRY [§241 6. Reduce the equations in exercises 1 and 2 to the projection form, the projecting planes being perpendicular to the xy and xz-planes. 7. Reduce the equations in exercises 3 and 4 to the projection form, the projecting planes being perpendicular to the xy and j/z-planes. 8. Reduce the equations in exercises 1-4 to the symmetric form. Find the equations in projection form of the straight lines in exercises 9-18. The projecting planes are to be taken perpendicular to the xy- and zz-planes whenever possible. 9. Passing through the points (3, —6, 4) and (—2, 5, 1). 10. Passing through the points (—2, 1, 2) and (3, — 1, 4). 11. Passing through the points (2, 1,-3) and (2, 3, —4). 12. Passing through the points (2, 5, 6) and (2, 5, 7). 13. Passing through the point (1, —3, 4) with direction cosines in the ratio 3 : — 1 : 2. 14. Passing through the point (3, —1, 2) and parallel to the r-axis. 15. Passing through the point (3, — 1, 2) and perpendicular to the z-axis. 16. Passing through the point (3, —1, 2) and parallel to the line of exercise 1. 17. Passing through the point (3, — 1, 2) and making right angles with the plane x — 2y + z = 3. 18. Passing through the origin and perpendicular to the lines x ~ X " ~2~ ~ ^2" and ^12 " "4 3~' 19. Find the cosine of the acute angle between the lines x-3 y + 2 z-6 jS-2 y-5 z -4- SB -=8- = -T- ,and ^ J""? 20. Find the cosine of the angle between the line 2x — 7y — 7z = —8, x — 2y — z = 5, and the line 12z — 15y — 2z = 70, 5x — by — z = 24. 21. Prove that the two lines x + y + 2 = 0, 2s — y + 3z = 7, and Zx + 4y + 2z = -3, -6s + 2y + lOz « meet in a point. 22. Prove that the planes 2x + 2y + z + 4 = 0, 4s+ j/ — z — 7=0, and 2x + Sy + 2z + 9 = 0, meet in a straight line and find its direction cosines. Find the equations of the planes that satisfy the conditions of exercises 23-26. 23. Passing through the point (2, 1, 3) and the line 3s + 5y-6z+9=0, 2x + 2y - 2z + 1 - 0. 24. Passing through the point ( — 1, —2, —3) and the line x — 1 y + 1 z + 5 3 2 3 Digitized by Google §241] SUMMARY OF FORMULAS 303 25. Passing through the parallel lines x - 1 y-3 2 + 1 A x-4 y-2 2+3 "I =7 6~ ,and -l =7- = ^6- 26. Passing through the intersecting lines x y + 5 2+4 , -7x - 1 y 7«-9 —} [ = "^- == ^9- ,and — 2 1--TT" 27. Find the equation of a plane through the line x + V - 2z + 2 - 0, 3s + 8j/ - 62 + 4 - 0, and perpendicular to the plane 7x + 2y + 22 — 10 = 0. 28. Find the equation of a line lying in the plane 2x — 2y + z + 11 - 0, passing through the point (—3, 2,-1), and parallel to the plane 2x + Zy - 42 + 5 - 0. SUMMARY OF FORMULAS [1] (1) OP* - OPx + PiiY (2) PxP 2 = OP 2 - OPi. [2] P1P2 = x% - *i. [2J P1P2 « s 2 - a? t , [2,] P1P2 « 1/2 - yi. [3] d = V(*i - x 2 ) 2 + (yi - y % )*. ri + r 2 ri + r 2 [6] m - tan a = yi "" y> - Xi — Xa [7] tan <p = y-j- -• 1 + mim 2 [8] For parallel lines, m\ = m 2 . [9] For perpendicular lines, mi = > and m 2 = W%2 Ml [10] x = p cos 0, y = p sin 0, x 2 + j/ 2 = p 2 . [11] p = V* 2 + y\ = tan- 1 J- [12] x = xf + h, y =y' + k. [12i] a/ - * - ft, 1/ = y -k. [13] x = x f cos <p — y' sin $>, y = x' sin tf> + j/' cos ^>. [13i] x' = x cos ^> + y sin ^, y 1 = y cos ^> — x sin ^>. Digitized by Google 304 ANALYTIC GEOMETRY [14] A = \{x x yt - 3*1/1 + x t y z - x z y t + x 9 y t - x x yt). [15] y - y x = m(x - Si). [16] y = mx + 6. [17] y-yi-J 1 -^ 1 («-»0. Xi — Xt [19] a; cos -f y sin tf — p =0. [20] Ax + By + C = 0. Ax By C roil _l y 4. = 1 J ±VaT+b*^ ±V3h-b» t ±VI i lTB i ' -c v ~ ±va* + b* _ Ax x + By, + C 1231 d_ ±VA* + 2** A t x + Bty + d A t x + B t y + C, lMi Va 1 * + b 1 * ± Vas + b,* [25] (x - ft)* + (V - k)* = r*. [26] x* + y* - r*. [27] x* + y* + Da; + Ey + F = 0. [28] y* - 2px. [29] x* = 2py. [30] (y - A)* - 2p(x - h). [30i] (x - *)• = 2p(y - k). w ' - r=V#* psiS+S-i- 6* M fe- ffi + St- fl! . L Digitized by Google SUMMARY OF FORMULAS [37]y = £x, y- -£*. 305 o* 6* [39] tan2p- j4^- [40] * = a(0 - sin 0), y = o(l - cob 0). x = (a - 6) cos + 6 cob ^ a ~ ^ 9, y - (a - 6) sin - b sin (a T 5 ? 0. o [42] x 1 + y» = o». x - (a + 6) cos - b cos ^4~^ »» [41] [43] y - (a + 6) sin e - b sin (o ± 6) g. f . f x = a cos + a $ sin 0, ^ y = a sin — a cos 0. [45] Axix + JBxiy + \Bxyi + Cyw + J2)x + \Dxi + hEy + \E yi + F = 0. ^Jy-y^H. <*-*>• ax I 'as = a?i [47] y - yi = - ^ (3 - «i). da: s = £i [«] d - V(*i- x 2 ) 2 + <yt - y 2 ) 2 + (zi - *,)■. 20 Digitized by Google 806 ANALYTIC GEOMETRY li9] *° r l + r t ' Vo ~ n + r, ' *° ~ n + r t [50 ]cos« = ?^- 1 , ^ fi^yJL^Jll, ^y = e -^p. [51] cos 2 a + cos 2 + cos 2 7 = 1. [52] p = ± y/x 1 + y* + z*, cos a = ±Vx 2 + 2/ 2 + * 2 y * cos = - — t=======, cos y = — x* + y* + z* ±Vx* + y* + z* [53] a: = p cos a, y = p cos 0, 2 = p cos 7. [54] x = p sin ?> cos B, y, = p sin <p sin 0, z = p cos <o. [54J p = ± V^+l/* + z\ 0= tan" 1 J' X = pit*-1 > £0 = COS -1 — — = sin -1 - — / . , > <p = cos" x* + y* ±Vx 2 + y 2 + 2 2 [55] cos = cos ai cos a 4 + cos p\ cos /3« + cos 71 cos 74. [56] (x-ft)*+(y-fc) J +(2-Z)* = r 2 . [57] Ax + By + Cz + D = 0. [58] x cos a + y cos /3 + z cos 7 = p. [59] cos a = , > cos = , — — > ±VA* + B* + C* ±VA* + B* + C* C -D COS "V = ; — V — , • ±VA* + B* + C* ±VA i + B* + C* [60] l + l + l-l. [6i] cos • = ^> + **. + m ^ + Ifo + C* + P ±VJ r +B 2 + C* [63l] iLZ* = 1LZ11! = L=JL». L cos a cos f$ cos 7 [63,] ^ = ^^ = i^i [64] *"** = V-Vi = £JZ*. a?j - Xi 2/2 — 2/1 «2 — *i Digitized by Google TABLES I. Four-place Table of Logarithms. II. Table of Natural and Logarithmic Sines, Cosines, Tangents, and Cotangents of Angles Differing by Ten Minutes. Digitized by Google 308 ANALYTIC GEOMETRY TABLE I.— COMMON LOGARITHMS V. 1 9 3 4 5 6 7 8 • 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 19 0702 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 1130 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 8010 3032 3054 3075 3096 3118 3139 8160 8181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 8444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 3802 3820 3838 8856 3874 3892 3909 3927 3945 3962 25 3979 8997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4698 4564 n 4594 4742 4609 20 4624 4639 4654 4669 4683 4713 4757 80 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 6038 n 4914 4928 4942 4955 4969 4983 4997 5011 5145 ftP 5051 5065 5079 5092 5105 5119 5132 5159 5172 83 84 6185 5315 5198 5328 5211 5340 5224 5358 5237 5366 6250 5378 6263 5276 6289 5391 5403 5418 5302 5428 85 5441 5453 5465 5478 5490 5502 5514 6*2t 5539 5551 86 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 88 5798 5809 5821 5832 5843 5855 5866 6877 5888 5899 30 5911 6922 5933 5944 6955 5966 5977 6988 6999 6010 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 40 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 K. 1 2 3 4 5 6 7 8 9. Digitized by Google TABLES TABLE I.— COMMON LOGARITHMS.— Contfmud 309 .-*. 1 8 4 5 7 8 9 n 7404 7412 7410 7497 7427 7435 7443 7451 7459 7466 7474 7482 7549 7490 7505 7513 7520 7528 7536 7543 7551 7627 07 7566 7574 7582 7589 7597 7004 7612 7619 8 88 7642 7640 7723 7657 7664 7672 7745 7679 7686 7694 7701 7716 7731 7738 7752 7760 7767 7774 00 7782 7780 7790 7803 7875 7810 7882 7818 7825 7882 7839 7846 01 7853 7860 7868 7889 7896 7903 7910 7917 8 7024 7931 7938 7945 7962 7959 7966 7973 7980 7987 7993 8000 8007 8014 8021 « 8035 8041 8048 8055 04 8002 8069 8070 8082 8089 8102 8109 8116 8122 00 8129 8130 8142 8149 8150 8162 8228 8160 8176 8182 8189 00 8195 8202 8209 8215 8222 8235 8241 8248 8254 07 8201 8267 8274 8280 8287 8293 8299 8306 8312 8319 00 00 8325 8388 8331 8395 8338 8401 8344 8407 8351 8414 8357 8420 8363 8420 8370 8432 8376 8439 8382 8445 70 71 8451 8513 8457 8519 8463 8525 8470 8531 8476 8537 8482 8543 8488 8549 8494 8555 8500 8561 8506 8567 7* 8573 8579 8639 8585 8591 8597 8603 8600 8615 8621 8627 73 8833 8645 8651 8657 8663 8669 8675 8681 8680 74 8692 8698 8704 8710 8710 8722 8727 8733 8739 8745 75 8751 8750 8769 8768 8774 8779 8785 8791 8797 8802 70 8808 8814 8820 8825 8831 8837 8842 8899 8848 8854 8859 77 8885 8921 8871 8876 8882 8887 8893 8904 8910 8915 78 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 00 9031 9080 9042 9047 9053 9058 9063 9069 9074 9079 01 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 89 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 88 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9370 9325 9330 9335 9340 88 9345 9350 9355 9360 9365 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 80 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 OQ 9542 9547 9595 9552 9557 9605 9562 9566 9571 9576 9581 9586 91 9590 9600. 9647 9609 9614 9619 9624 9628 9633 00 9638 9643 9652 9657 9661 9666 9671 9675 9680 Si 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 g 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 9823 9827 9832 9877 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9881 9886 9890 9894 9899 9903 9908 S3 9912 9956 9917 9921 9926 9930 9934 9939 9943 9948 9952 9961 9965 9969 9974 9978 9983 9987 9991 9996 V. 1 3 4 5 7 8 9 Digitized by Google 310 ANALYTIC GEOMETRY TABLE IL— TRIGONOMETRIC FUNCTIONS Angles Sines Cosines Tangent* Cotangents Angles Nat. Loc Nat. I*c 1 0000 0.0000 Nat. Loc Nat. Log. (TOT .0000 oo .0000 00 00 ' 00 90° 00* 10 .0029 7.4637 1.0000 0000 .0029 7.4637 343.77 2.5363 50 20 .0058 76481.0000 0000 .0058 7648 171.89 2352 40 30 .0087 9408 1.0000 0000 .0087 9409 114.59 0591 30 40 .0116 8.0658 .9999 0000 .0116 8.0658 85.940 1.9342 20 50 .0145 1627 .9999 0000 .0145 1627 68.750 8373 10 1°<& .0175 8.2419 •9998 9.9999 .0178 8.2419 57.290 1.7581 89° 00' 10 .0204 3088 .9998 9999 .0204 3089 49.104 6911 50 20 .0233 3668 .9997 9999 .0233 3669 42.964 6331 40 30 .0262 4179 .9997 9999 .0262 4181 38.188 5819 30 40 .0291 4637 .9996 WJtfo .0291 4638 34.368 5362 20 50 .0320 5050 .9995 9998 .0320 5053 31.242 4947 10 2*00' .0349 8.5428 .9994 9.9997 .0349 8.5431 28.636 1.4569 88° (W 10 .0378 5776 .9993 9997 .0378 5779 26.432 4221 50 20 .0407 6097 .9992 9996 .9990 9996 .0407 6101 24.542 3899 40 30 .0436 6397 .0437 6401 22.904 3599 30 40 .0465 6677 .9989 9995 .0466 6682 21.470 3318 20 50 .0494 6940 .9988 9995 .0495 6945 20.206 3055 10 3°00' .0523 8.7188 .9986 9.9994 .0524 8.7194 19.081 1.2806 87°00 f 10 .0552 7423 .9985 9993 .0553 7429 18.075 2571 50 20 .0581 7645 .9983 9993 .0582 7652 17.169 2348 40 30 .0610 7857 .9981 9992 .0612 7865 16.350 2135 30 40 .0640 8059 .9980 9991 .0641 8067 15.605 1933 20 50 .0669 8251 .9978 9990 .0670 8261 14.924 1739 10 4°00 f .0698 8.8436 .9976 9.9989 .0699 8.8446 14.301 1.1554 86° (W 10 .0727 8613 .9974 9989 .0729 8624 13.727 1376 50 20 .0756 8783 .9971 9988 .0758 8795 13.197 1205 40 30 .0785 8946 .9969 9987 .0787 8960 12.706 1040 30 40 .0814 9104 .9967 9986 .0816 9118 12.251 0882 20 50 .0843 9256 .9964 9985 .0846 9272 11.826 0728 10 5°00 f .0872 8.9403 .9962 9.9983 .0875 8.9420 11.430 1.0580 85° 0C 10 .0901 9545 .9959 9982 .0904 9563 11.059 0437 50 20 .0929 9682 .9957 9981 .0934 9701 10.712 0299 40 30 .0958 9816 .9954 9980 .0963 9836 10.385 0164 30 40 .0987 9945 .9951 9979 .0992 9966 10.078 0034 20 50 .1016 9.0070 .9948 9977 .1022 9.0093 9.7882 0.9907 10 6° 00* .1045 9.0192 .9945 9.9976 .1051 9.0216 9.5144 0.9784 84° 00* 10 .1074 0311 .9942 9975 .1080 0336 9.2553 9664 50 20 .1103 0426 .9939 9973 .1110 0453 9.0098 9547 40 30 .1132 0539 .9936 9972 .1139 0567 8.7769 9433 30 40 .1161 0648 .9932 9971 .1169 0678 8.5555 9322 20 50 .1100 0755 .9929 9969 .1198 0786 8.3450 9214 10 7° 00* .1219 9.0859 .9925 9.9968 .1228 9.0891 8.1443 0.9109 83° 00* 10 .1248 0961 .9922 9966 .1257 0995 7.9530 9005 50 20 .1276 1060 .9918 9964 .1287 1096 7.7704 8904 40 80 .1305 1157 .9914 9963 .1317 1194 7.5958 8806 30 40 .1334 1252 .9911 9961 .1346 1291 7.4287 8709 20 50 .1363 1345 .9907 9959 .1376 1385 7.2687 8615 10 8° 00, .1392 9.1436 .9903 9.9958 .1405 9;1478 7.1154 0.8522 82*00' 10 .1421 1525 .9899 9956 .1435 1569 6.9682 8431 50 20 .1449 1612 .9894 9954 .1465 1658 6.8269 8342 40 30 .1478 1697 .9890 9952 .1495 1745 6.6912 8255 30 40 .1507 1781 .9886 9950 .1524 1831 6.5606 8169 20 50 .1536 1863 .9881 9948 .1554 1915 6.4348 8085 10 9 # 00' .1564 9.1943 .9877 9.9946 .1584 9.1997 6.3138 0.8003 81° 00' Nat. Log. Nat. Log. Nat. Log. Nat. Log. Angle* Cosines Sines Cotangents Tangents Angles Digitized by Google TABLES 311 TABLE IL— TRIGONOMETRIC FUNCTIONS— ConHnutd Angles Sines Cosines , Tangents Cotangents Angles VW Nat. Log. .1564 0.1943 Nat. Log. .9877 9T9946 Nat. Log. .1584 971997 Nat. Log. 6.3138 0.8003 81° Off 10 .1503 2022 .9872 9944 .1614 2078 6.1970 7922 50 20 .1622 2100 .9868 9942 .1644 2158 6.0844 7842 40 90 .1650 2176 .9863 9940 .1673 2236 5.9758 7764 30 40 .1679 2251 .9858 9938 .1703 2313 5.8708 7687 20 . 50 .1706 2324 .9853 9936 .1733 2389 6.7694 7611 10 10* w .1736 9.2397 .9848 9.9934 .1763 9.2463 5.6713 0.7537 80 # 00' 10 .1765 2468 .9843 9931 .1793 2536 5.5764 7464 50 £ .1794 2538 .9838 9929 .1823 2609 5.4845 7391 40 .1822 2606 .9833 9927 .1853 2680 5.3955 7320 30 40 .1851 2674 .9827 9924 .1883 2750 5.3093 7250 20 50 .1880 2740 .9822 9922 .1914 2819 5.2257 7181 10 ii° <xr .1908 9.2806 .9816 9.9919 .1944 9.2887 5.1446 0.7113 79° 00* 10 .1937 2870 .9811 9917 .1974 2953 5.0658 7047 50 20 .1965 2934 .9805 9914 .2004 3020 4.9894 6980 40 90 .1994 2997 .9799 9912 .2035 3085 4.9152 6915 30 40 .2022 3058 .9793 9909 .2065 3149 4.8430 6851 20 so .2051 3119 .9787 9907 .2095 8212 4.7729 6788 10 12° W .2079 9.3179 .9781 9.9904 .2126 9.3275 4.7046 0.6725 78*00' 10 .2108 3238 .9775 9901 .2156 8336 4.6382 6664 50 20 .2136 3296 .9769 9899 .2186 3397 4.5736 6603 40 30 .2164 3353 .9763 9896 .2217 3458 4.5107 6542 30 40 .2193 3410 .9757 9893 .2247 3517 4.4494 6483 20 m , JBKM £416 jum mm ~£2» ISM 4.JOT *CM 10 13* (W .2250 9 .3521 .*744 9.9887 .2309 9.3634 4.3315 0.6366 rrw 10 .2278 3575 .9737 9884 .2339 3691 4.2747 6309 so 20 .2306 3629 .9730 9881 .2370 3748 4.2193 6252 40 30 .2334 3682 .9724 9878 .2401 3804 4.1653 6196 30 40 .2363 3734 .9717 9875 .2432 3859 4.1126 6141 20 50 .2391 3786 .9710 9872 .2462 3914 4.0611 6086 10 14° W .2419 9.3837 .9703 9.9869 .2493 9.3968 4.0108 0.6Q32 76° 00* 10 .2447 3887 .9690 9866 .2524 4021 3.9617 5979 50 20 .2476 3937 .9689 9863 .2555 4074 3.9136 5926 40 30 .2504 3986 .9681 9859 .2586 4127 3.8667 5873 30 40 .2532 4035 .9674 9856 .2617 4178 3.8208 5822 20 50 .2560 4083 .9667 9853 .2648 4230 3.7760 5770 10 15° W .2588 9.4130 .9659 9.9849 .2679 9.4281 3.7321 0.5719 75° (XT 10 .2616 4177 .9652 9846 .2711 4331 3.6891 5669 50 20 .2644 4223 .9644 9843 .2742 4381 3.6470 5619 40 30 .2672 4269 .9636 9839 .2773 4430 3.6059 5570 30 40 .2700 4314 .9628 9836 .2805 4479 3.5656 5521 20 50 .2728 4359 .9621 9832 .2836 4527 3.5261 5473 10 16° <xr .2756 9.4403 .9613 9.9828 .2867 9.4575 3.4874 0.5425 74° 00* 10 .2784 4447 .9605 9825 .2899 4622 3.4495. 5378 50 20 .2812 4491 .9596 9821 .2931 4669 3.4124 5331 40 30 .2840 4533 .9588 9817 .2962 4716 3.3759 5284 30 40 .2868 4576 .9580 9814 .2994 4762 3.3402 5238 20 50 .2896 4618 .9572 9810 .3026 4808 3.3052 5192 10 17° W .2924 9.4659 .9563 9.9806 .3057 9.4853 3.2709 0.5147 73° 00* 10 .2952 4700 .9555 9802 .3089 4898 3.2371 5102 50 20 .2979 4741 .9546 9798 .3121 4943 3.2041 5057 40 v 30 .3007 4781 .9537 9794 .3153 4987 3.1716 5013 30 40 .3035 4821 .9528 9790 .3185 5031 3.1397 4969 20 50 .3062 4861 .9520 9786 .8217 5075 3.1084 4925 10 18 # 0(K .3090 0.4900 .9511 9.9782 .3249 9.5118 3.0777 0.4882 72° W. Nat. Log. Nat. Log. Nat. Log. Nat. Log. Angle. Cosines Sines Cotangents Tangents Angles Digitized by Google 312 ANALYTIC GEOMETRY TABIA XL-TWOONOMBTRKJ FUNOITONS-CoitfftNM* 90 80 40 50 "•ff 90 80 40 50 20 80 40 50 *'V 20 80 40 50 22° 00* 10 20 80 40 50 33° 0C 10 20 80 40 50 24° aor 10 20 y 30 40 50 20 30 40 50 20 80 40 50 WW 3907 0.5919 3934 5948 3901 5978 3987 6007 4014 or 4041 6005 4067 9.0093 4094 6121 4120 6149 4147 6177 4173 6205 4200 6T~ 4326 9.6259 4253 6286 4279 6313 4305 6340 4331 6366 4358 6392 4384 9.6418 4410 6444 4436 6470 4462 6495 4488 6521 4514 6546 4540 9.6570 Nat. Log. Coynes Nat. .9455 ,9446 .9436 .9426 .9417 .9407 .9397' .9387 .9377 .9367 .9356 .9346 .was .9325 .9315 .9304 .9293 .9239 .9076 « .9063 9.9573 ,9051 9567 .9038 9561 .9026 9555 .9013 9549 ,9001 9543 .8988 9.9537 .8975 9530 .8962 9524 .8949 9518 .8936 9512 .8923 9505 .8910 9.9499 Nat. Log. Sines TanffenU Cotangent* Nat. Log 8.0777 0748L 3.0475 4839 *3' _ fc ._ ,3541 ^491 .3574 5531 ,3607 6571 9.5611 3.0178 2.9887 2.9600 12.9319 19.9042 0.4630 2.8770 4589 8502 4549 _.&239 4509 2.7980 4469 3.7725 4429 8,7475 0.4389 2.7228 4350 2.6985 9.6746 2.6511 2,627$ 2.6051 0.4158 2.5826 41 2.5605 2.5386 2.5172 2.4960 or .4314 .4348 .4383 .4417 .4040 9.6064 .4074 6100 .4108 6136 6172 6208 6243 9.6279 6314 6348 6383 6417 6452 .4452 0.6486 .4487 6520 .4522 6553 .4557 6587 .4592 6620 .462a 6654 .4663 9.6687 .4699 6720 .4734 6752 .4770 6785 .4806 6817 .4841 6850 .4877 9.6882 .4913 6914 .4950 6946 .4986 6977 .5022 7009 .5059 7040 .5095 9.7072 Nat. Log. Cotangent! 4046 4009 3972 2.4751 0.3936 2.4545 3900 3864 4797 4755 4713 4671 4311 4273 4234 4196 ,.31_ 2.2998 2.2817 2.2637 2.2460 0.3514 2.2286 3480 3447 3413 2.2113 2.1943 2.1775 „„ 2.1609 3346 2.1445 0.3313 2.1283 3880 2.1123 3248 2.0965 3215 2.0809 3183 2.0655 3150 2.0503 0.3118 2.0353 3086 2.0204 2.0057 1.9912 1.9768 3054 3023 2991 2960 1.9626 0.2928 Nat. Log. Tangents 40 30 20 10 40 30 20 10 70°00 r 50 40 30 20 10 40*00? 50 40 30 20 10 68° 00* 50 40 30 20 10 67° 00* 50 40 30 20 10 66° 00? 50 40 30 20 10 65° 00? 50 40 30 20 10 64*00? 50 40 30* 20 10 63* c*y Digitized by Google , TABLES 313 TABUI n.—TRiaONOMETRIO TmiCnom-ConHmu* A*gi* Ones Cosines Tangents Cotangents Aaglss iroff .*W 076&0 %*• Loft .8910 979499 Nat. Log. .6095 9.7072 Nat. Lag. 1.9626 0L2928 68*0ff 10 .4566 6595 .8897 9492 .5132 7103 1.9486 2897 50 20 .4592 6620 .4617 6644 .8884 9486 .5169 7134 1.9347 2866 40 80 .8870 9479 .5206 7165 1.9210 2885 80 40 .4648 6668 .4669 6602 .8857 9473 .6243 7196 1.9074 2804 1.8940 2774 20 50 .8848 9466 .5280 7226 10 «raor .4698 0.6716 .4720 6740 .8820 0.9459 .5817 0.7257 1.8807 0.2743 62* OOP 10 .8816 9453 .5354 7287 1.8676 2713 50 20 .4746 6768 .8802 9446 .5392 7317 1.8546 2Q83 40 80 .4772 6787 .4797 6810 .8788 9439 ..5430 7348 1 .8418 2652 30 40 .8774 9432 .8760 9425 .5467 7378 1 .8291 2622 20 50 .4828 6888 .5505 7408 1.8165 2592 10 iroff .4848 0.6856 .8746 0.9418 .8548 0.7488 1.8040 0.2562 1.7917 2533 61° Off 10 .4874 6878 .8732 9411 .5581 7467 50 20 .4899 6901 .8718 9404 .5619 7497 1.7796 2503 40 80 .4924 6923 .8704 9397 .5658 7526 1.7676 2474 30 40 .4950 6946 .8689 9390 .8676 9383 .5696 7556 1.7556 2444 20 50 .4975 6068 .5735 7585 1.7437 2415 10 *V .5000 0.69M .8660 0.0375 .5774 0.7614 1.7821 0.2386 60° Off .5025 7012 .8646 9368 .6812 7644 1.7205 2356 50 20 .5050 7033 .8631 9361 .5851 7673 1.7090 2327 40 80 .5076 7055 .8616 9353 .5890 7701 1.6977 2299 30 40 .5100 7076 .8601 9346 .6930 7730 1 .6864 2270 20 50 .5128 7097 .8587 9338 .5960 7769 1.6753 2341 10 81° Off .5180 0.7118 .5175 7139 .8572 0.9331 .6009 0.7788 1 .6643 0.2212 t.6534 2184 59° Off 10 .8557 9323 .6048 7816 50 20 .5200 7160 .8542 9315 .6088 7845 1.6426 2165 40 80 .5225 7181 .8526 9308 .6128 7873 1.6319 2127 30 40 .6250 7201 .8511 9300 .6168 7902 1.6212 2098 20 50 .5276 7222 .8496 9292 .6208 7930 1.6107 2070 10 Wtt' .8290 0.7242 .5324 7262 .8480 0.9284 .6240 0.7958 1.6003 0.2042 58° Off 10 .8465 9276 .6289 7986 1.5900 2014 50 20 .5348 7282 .5378 7302 .8450 9268 .6330 8014 1.5798 1986 40 80 .8434 9260 .8418 9252 .6371 8042 1.5697 1958 30 40 .5398 7322 .6412 8070 1.5597 1930 20 50 .5422 7342 .8403 9244 .6453 8097 1.5497 1903 10 W°0ff .8446 9.7361 .8387 0.9236 .6494 9.8125 1.5399 0.1875 57° Off 10 .5471 7380 .8371 9228 6536 £153 1.5301 1847 50 20 .5495 7400 .8355 9219 .6577 8180 1 . 5204 1820 40 30 .5519 7419 .8339 9211 .8323 9203 .6619 8208 1.5108 1792 30 40 .5544 7438 .6661 8235 1 .5013 1765 20 50 .5568 7457 .8307 9194 .6703 8263 1.4910 1737 10 34° W .5592 9.7476 .8290 0.9186 .6745 0.8290 1.4826 0.1710 56° Off 10 .5616 7494 .8274 9177 .6787 8317 1.4733 1683 50 20 .5640 7513 .8258 9169 .6830 8344 1.4641 1656 1.4550 1629 40 80 .5664 7531 .8241 9160 .6873 8371 30 40 .5688 7550 .8225 9151 .6916 8398 1.4460 1602 20 50 .5712 7568 .8208 9142 .6950 8425 1.4370 1575 10 WOO* .5736 9.7586 .8192 0.9134 .7002 0.8452 1.4281 0.1548 55° Off 10 .5760 7604 .8175 9125 .7046 8479 1.4193 1521 50 20 .5788 7622 .8158 9116 .7089 8506 1.4106 1494 40 30 .5807 7640 .8141 9107 .7133 8533 1.4019 1467 30 40 .5831 7657 .8124 9098 .7177 8559 1.3934 1441 20 50 .5854 7675 .8107 9089 .7221 8586 1.3848 1414 10 86° 00* .5878 9.7692 .8090 9.9080 .7265 0.8613 1.3764 0.1387 54° Off Nat. Log. Nat. Log. Nat. Log. Nat. Log. Angles Cosines Sines Cotangents Tangents Angles Digitized by Google 314 ANALYTIC GEOMETRY TABLE II,— TRIGONOMETRIC FUNCTIONS— & Angles Bnm finsinw Tangents 1 Cotangents Angles M»OK Nat. Log. .1878 9/7602 Nat. Log. .8000 07*380 Nat. Log. Nat. Log. .7266 9.8613 1.3764 0.1387 54? vr 10 .ajpl 7710 .8073 0070 .7310 86301.3680 1361 50 90 .5095 7727 .8066 0061 .7355 86661.3507 1334 40 30 .5048 7744 .8030 0052 .7400 86021.3514 1308 30 40 .5072 7761 .8021 0042 .7445 8718,1.3432 1282 90 60 .5006 7778 .8004 0033 •7400 87451.3351 1255 10 troo' .0018 0.7705 .7086 0.0023 •7536 0.877111.3270 0.1220 53° 00* 10 .6041 7811 .7000 0014 .7581 8707 1.3100 1203 50 90 .6066 7828 .7051 0004 .7627 88241.3111 1176 40 30 .6068 7844 .7034 8005 .7673 88501.3032 1150 30 40 .6111 7861 .7016 8085 .7790 88761.2054 1124 90 50 .6134 7877 .7806 8075 .7766 800211.2876 1008 10 3T0CT .6157 0.7803 .7880 0.8065 .7818 0.80281.2700 0.1072 52° OCT 10 .6180 7010 .7862 8055 .7860 89541.2723 1046 50 90 .6209 7026 .7844 8045 .7007 8080,1 .2647 1020 40 30 .6225 7041 .7826 8035 .7054 0006 1 2579 0004 30 40 .6248 7057 .7808 8025 .8002 00321.2407 0068 90 50 •6271 7073 .7700 8015 .8050 0058 1.9423 0042 10 wrw .6208 0.7080 .7771 0.8005 .8006 0.0064 1.2340 0.0016 51*00' 10 .6316 8004 .7753 8805 .8146 0110 1.2276 0800 50 90 .6338 8020 .7735 8884 .8105 0135 1.2203 0865 40 30 .6361 8035 .7716 8874 .8243 0161 1.2131 0830 30 40 .6383 8050 .7608 8864 .8202 0187 1.2050 0613 90 50 .6406 8066 .7670 8853 .8342 9212 1.1088 0788 10 lO'OO' .6428 0.8081 .7660 0.8843 .8301 0.0238 1.1018 0.0762 5O»0O' 10 .6450 8006 .7642 8832 .8441 0264 1.1847 0736 50 90 .6472 8111 .7623 8821 .8401 0280 1.1778 0711 40 30 .6404 8125 .7604 8810 .8541 0315 1.1706 0685 30 40 .6517 8140 .7585 8800 .8501 9341 1.1640 0650 20 50 .6530 8155 .7566 8780 .8642 • 0366 1.1571 0634 10 41° <xr .6561 0.8160 .7547 0.8778 .8603 0.0302 1.1504 0.0608 49° 00* 10 .6583 8184 .7528 8767 .8744 0417 1.1436 0583 50 90 .6604 8108 .7500 8756 .8796 0443 1.1360 0557 40 30 .6626 8213 .7400 8745 .8847 04681.1303 0532 30 40 .6648 8227 .7470 8733 .8899 0494 1.1237 0606 20 50 .6670 8241 .7451 8722 .8952 0519 1.1171 0481 10 49° W .6601 0.8255 .7431 0.8711 .0004 9.9544 1.1106 0.0456 48° W 10 .6713 8260 ,7412 8600 .9057 9570 1.1041 0430 50 20 .6734 8283 .7302 8688 .9110 9595 1.0977 0405 40 30 .6756 8207 .7373 8676 .9163 9621 1.0913 0379 1.0850 0354 30 40 .6777 8311 .7353 8665 .9217 9646 20 50 .6700 8324 .7333 8653 .9271 9671 1.0786 0329 10 43° 00* .6820 0.8338 .7314 0.8641 .9325 0.9697 1.0724 0.0303 47°O0T 10 .6841 8351 .7204 8620 .9380 9722 1.0661 0278 50 90 .6862 8365 .7274 8618 .9435 9747 1.0599 0253 40 30 .6884 8378 .7254 8606 .9490 9772 1.0538 0228 30 40 .6005 8301 .7234 8504 .9545 9798 1.0477 0202 20 50 .6026 8405 .7214 8582 .9601 9823 1.0416 0177 10 44 # 00' .6047 0.8418 .7103 0.8560 .9657 9.0848 1.0355 0.0152 46°O0T 10 .6067 8431 .7173 8557 .0713 0874 1 .0295 0126 50 20 .6088 8444 .7153 8545 .0770 9899 1.0235 0101 40 30 .7000 8457 .7133 8532 .9827 9924 1.0176 0076 30 40 .7030 8460 .7112 8520 .9884 9949 1.0117 0051 20 50 .7050 8482 .7002 8507 .9942 9975 1 .0058 0025 10 ^•OCK .7071 0.8405 .7071 0.8405 1.0000 0.0000 1.0000 0.0000 45° Off Nat. Log. Nat. Log. Nat. Log. Nat. Log. Angles Cosines 'Sines Cotangents Tangents Angles Digitized by Google ANSWERS » Page 12. Art. 12. I. 1, 5, -8, -10, -3, 11, -16, -13. Pages 1*, 16. Art. 17. 10. (5\/2, 5\/2), (10V2, 0), (5\/2, - 5V2). II. x - y = 0, x + ? - 0, 2x - y - 0. 12. (0,0), (a,0), (}a, ±W§). 18. (8, 0), (4, 4V3), (-4,4VU, (-8, 0), (- 4,-4\/3), (4, -4\/3). Pages 18, 19. Art. 19. 1. (1) 15, (2) 18.385-, (3) 13.153-, (4) 16.279-. 5. (1) 8.602+, 8.062+, 12.369+. (2) 11.402-, 8.062+, 8.062+. 7. (3, -2) or (3, 14). 8. (1, 3). 9. (-1, 3), (-3, 5), or (13, -1). 10. (5 + 4\/3, 6 + 3V3) or (5 - 4a/3, 6 - 3\/3). 11. x 1 + y* - 6s - $y - 0. 12. 5x - 7y - 26 - 0. 15. 7.550-. Page 20. Art. 20. 1. 2f units to the right of P u 12 units to the left of Pi. 2. Division point between two points and 3 in. from first. Division point beyond second point, 5 in. from first. Pages 23, 24. Art 22. 1. (-2, 1). 2. (li, 3i). 3. (3, H). 4. (-22, 14). 6. (i, -J). 7. (i, J), (-24, 28). 8. (1J, -1), (-li, -2i;, (-1, -4J,. 9. (2f, 3*). 10. 10.050-, 11.180+, 12.806+. 13. (11, 14). 14. (-1,0), (-4, -2). Pages 28, 29. Art. 28. 1. (1) 1, (2) -1, (3) 1.732, (4) 0.1010, (5) £-=-£ (6) -3.1463. 2. (1) 45°, (2) 135°, (3) 60°, (4) 5° 46', (5) tan~* £-=^ (6)107° 38'. 3. -*. 5. 5J. 7. 6. 8. 3x - 2y - 2 » 0. 9, If. 315 Digitized by Google 316 ANALYTIC GEOMETRY 10. &r - y + 12 - 0. 11. x + 2y - 11 = 0, (-3, 7). 12. 60° 15'. 18. 2.375. 14. 86° 11'. 15. lOf. 16. 3.732. 19. (lft, 3ft). 20. 0.6584. 21. 74° 56'. Pages 82, 83. Art 80. 1. (3\/2, 3V2,) (-V2,- V2), (-1, -|V3), (3^3,3), (-t,fV3),(-4 % /2 f 4 V ^2;,(-2, 0), (0,-6). 2. (1.532, 1.286), (1.026, 2.819), (5.629, -3.250), (-0.7714, 0.9192), (4.078, -1.902). 3. (8, 60°), (-8, 240°); (3V2, 225°), (-3<\/2, 45°); (V34, $9 Q 2')» (- V§4, 239° 2'); (2V2, 120°), (-2V5, 300°). f. 4.58. Pages 35, 36. Art 88. 2. (f\/2, JV2),(iV2, |\/2), (-5V2, 0),(3\/2,-4\/2). 3. (0, 0), (9, 2), (5, 11). 4. (7, 8). 5. 60°; (0, 0), (4, 0), (2 Wl. V^8 + %). Pages 38, 8*. Art 85. 1. (1) 76, (2) 31, (3) 200}, (4) 10. 2. 160. 8. 18. «. 72. . 7. i\pu>t sin (0i - Ox) + pips sin (0* - t ) + a»pi 8in fo — #i)]. 8. 98.29. Page 41. Art. 86. 1. (-34,2). Pages 41-43. General Exercises, 2. (0, 0), (8, 0), (0, 10), (^8, 10). 3. (-8, 0), (0, 0), (8, 10), (0, 10). 4. (4, 0), (0, 4), (-4, 0), (0, -4). 5. (a + iaVs, Ja), (}aV3,|a). 6. (0, -*a\/3), (}a, *a\/3), (-*<*, |aV3) or (0, *aV3), (ia,-i«V3), (-K -iaV3). 7. 4.799. 8. 2:5. 9. (6, -4), (14, -20). 10. (0, 9), (3, 0). 11. (1, 4). 12. «, V). 14. (7, 0) or (-2, 0). 15. 6 or -2. 16. (5, 0), (0, -1). 17. (5, 0), (-3, 4). 19. -}. 2a -^1.128. 21. -0.145. 22. 1.2337. 23. 2.25. 24. (0, 0), (4ft, lit), (0, 13), (-4ft, lift). 25. (0, 0), (5,0), (4, 3), (-1, 3). Page 49. Art. 44. 1. 5, -3*. 2. ±6, ±6. 3. ±4, ±8. 4. 1 ± iVZ, 1. i. 0, a 6. 0, 2 and 0. 7. -2, 1, 3 and ±\/6. 8. None. Digitized by Google ANSWERS 317 *agtl4. Art. 43. JL *V -* 3* - 0. 8. *» — 4*? + 4y* - 9 - 0. 3. *» - 8** + 15* - 0. 4. *• - x*y + *y* - y« - 16* + 16* - 0. 5. x«y - te 1 - 4*y + *y* - 6y* + 24 » 0. Page 86. Art. 49. I, (2V2, — 2V5), (— 2\/2,2x/2). 2. (t\/2, JVTi), (J\/2, -J\/l4). (-|V^,iVl4),(-!v^, WTi). 8. (24fc -12?), (3, 2). 4. (fVlO, lVT5,, (fVlO,- fVifi), (- JVT6, *Vl6), (-jVlO, -*\/l5). 5. (3, 4), (3, -4). 6. 4^2. Pages 67, 68. Art. 61. II. 22* + 120 - 1 - 0. 12. ** + y» + 12* + 16y «0. 13. *» + y % - 6* - Sy - 0. 14. ** + y» + 2* + 4y - 20 * 0. 16. 18* + 20y + 51 - 0. 16. 4* + 3y - 25 - 0, 2* - 6? + 29 - 0, 3* - y + 2 - 0. 17. 4* - y « 0. 18. 8** + 8y» + 112* - 30y + 347 - 0. 19. 2* - 3y + 24 - 0. 20. 7*» + 16y» - 112 - 0. 21. 5** - 4y» - 20 - 0. 22. 2* - 6* - 5 - 0, 2* - 6y - 15 - 0. 23. xy + 7* + 8y - 4 - 0. Page 61. Art. 64. 1. * - 2y - 8 = 0. 2. * + y + 6 - 0. 3. \/3* + y - VS - 5 - 0. 4. 3* - by + 13 - 0. 6. 5* + 3y - 1 - 0. 6. * - 3 - 0. 7. y - 4 ' - 0. 8. * - 2y + 5 = 0. 9. 3* - 4y - 11 - 0; 3* + 4y + 5 - 0. 10. 12* - by - 26 = 0. It. 11* - y - 16 - 0. 12. 7* - y - 5 - 0. Page 62. Art. 66. 1. * - 2y + 6 - 0. 2. 3* - y - 2 = 0. 3. 2* - 3y + 1 =* 0; 2* + 3y - 1 - 0. 4. 2* + 3y - 20 = 0. 6. * + y + 1 - 0. 6. * - 3y - 5 =* 0. Page 64. Art. 68. 1. 2* - Sy - 6 - 0. 2. 6* ' - y + 6 ~ 0. 3. 4* + 3y - 2 - 0. 4. 3* + 4y + 1 - 0. 6. * + V3y -6=0. 6. * + y/Zy + 6=0. 7. * - y + 2 V2 =0. 8. \/Sx + y + 2 - 0. 9. \/3* - y + 8 - 0. 10. V3* - y - 4 - 0. Digitized by Google 318 ANALYTIC GEOMETRY Pages 67, 68. Art 62. 1. -f, 2. 2. |, |. 8. |, 3. 4. 1, }. 5, *. 6. 2. 7. 2. 8. 3/ 9.?^?. 10. !b£I?. 11. scosO° + ysinO° + i «0,i. 5 5 12. s cos 90° + y sin 90° - | - 0, |. 18. 3s - 4y + 5 = 0. 14. x + 2y + 4 - 0. 15. x - 2y - 2 - 0. 16. * + y + 2 = 0. Page 69. Art 68. 1. f . 2. 1. 8. 1.4142. 4. 0.232. 8. 1.828. 6. 4.427. 7. 2, Jf, ft. 8. 5.233, 6.871, 3.757. Page 71. Art. 64. 1. x - Zy - 3 = 0. 2. 2s + 6y - 7 « 0. 3. x - 7y + 42 - 0. 4. 16s - 4y - 17 - 0. 5. 30s + lOy + 9 - 0. 6. 2x + y - 2 - 0. 7. 7s - 9y - 0, 4s + 6y - 21 - 0, 5s + y - 14 - 0. 8. x + Zy - 4 = 0, x - 7y - 19 = 0, 2s - 17 - 0. Page 72. Art. 65. I. mx - y + 2m + 3 = 0. 2. y - mx = 0. 8. ma; - y - Zm + 4 - 0. 4. | + | - 1. 5. mx — y — 4 = 0. 6. s cos -f y sin -r- 3 = 0. 7. s cos -f- 2/ sin - 7 « 0. 8. 2s -f- y - b = 0. 9. 3s + y — 6 — 0. 10. s + y — 6 =* or y = »w. Pages 74, 75. Art. 66. 1. 3s + 4y ± 6 = 0. 2. s - 2y - 0, s + y - 6 - 0. 8. 3s + 4y + 5 - 0, 5s + I2y - 13 - 0. 4. y - 2 - 0, 4s - Zy - 10 = 0. 5. 2s - y - 6 - 0. 6. 2s + y - 6 - 0. 7. s + 4y - 4 - 0. 8. s - y - 3 = 0. 9. s + y - 4 - 0, 3s + y - 6 - 0. 10. s + 3y - 6 = 0. II. s + VZy -6=0. 12. 3s - 4y ± 10 = 0. 13. 4s - Zy + 15 - 0, 4s + Zy - 15 - 0. 14. 3s + 4y ± 10 - 0, 4s + Zy ± 10 - 0. 15. 2s + y - 4 - 0. 16. 2s - (3 - 2y/2)y + 4 - 4\/2 = 0, 2s - (3 + 2V2)y + 4 + 4\/2 - 0. Digitized by Google ANSWERS 319 Page 77. Art. 67. 1. 4x - by + 1 = 0. 2. 13s + \2y - 62 - 0. 8. x - 2y + 6 - 0. 4. 22a; + lly - 14 - 0. 5. 5a? + I5y - 34 - 0. Page 78. Art. 68. 1. x*y — y* - xy + y* = 0. 2. *V ~ 3s*y - 3xy* + 2x» + 9xy + 2y» - 6a: - 6y + 4 - 0. Page 79. Art. 69. 1. p cos (0 - 45°) =3. 2. p cos (0 - 60°) - -2. 3. p sin — 7. 4. p cos 6 = —4. 5. p cos (0 - 135°) = -4. 6. p cos (0 - 315°) - 3. 7. p cos (0 - 45°) = J V2. 8. p cos = 3. 9. p sin - -7. 10. p cos (0 - 30°) = 2. 11. p cos (0 - 300°) = -3. 12. tan 0=2. 13. x - 3 - 0. 14. y - 4 - 0. 15. y - 6a: - 0. 16. x + y - 2 = 0. 17. a: + y - 3 = 0. 18. 4a; - 6y - 3 - 0. 19. 3a: ± 4y = 0. 20. 12a: ± 5y = 0. 21. a: - y + 2 = 0. 22. x - VSy -6=0. Pages 81-85. General Exercises. 2. (1) 2a: - Zy - 17 = 0. 3. (1) x - 7y - 33 = 0. (2) 7x - 2y - 21 = 0. (2) y = mx + k - mft. 4. (1) i. (2) 0.5883. (3) Vj_ 5. f i-v/82, fi VSI, V V5. _ 6. AVIS, H Vl3, if Vl3, A Vl3, A Vis. 7. (1) 7. 8. (1) a: + Sy - 58 = 0. (2) -f. (2) 21a: - 6y - 58 = 0. (3) 12a: + 9y - 116 = 0. (4) 15a: + 6y - 110 = 0. 9. 3a: - y - 1 = 0, 7a; + by - 15 = 0, x - 4y + 6 = 0, (H, H). 12. a; + V3y -4=0. 13. (1) x - y + 2 = 0. (2) 2a: - y - 1 = 0. (3) V3a: + y - 8 + 2 V3 = 0. 15. 2a; - jy + 6 = 0. 16. («, J). 17. a;' - by'+ 4\/2 = 0. 19. 4a; - by + 1 - 0, 7a: - Sy - 27 = 0, 3a: + 2y - 5 - 0. Digitized by Google 320 ANALYTIC GEOMETRY 20. s - y - 1 - 0. 21. s + y - 8\/2« 0. 22. 3* + 4y - 20 - 0. 23. 7x - 4y - 2 - 0. 24. x - 3y + 7 - 0, 13* + 9y - 5 - 0. 25. x - 2y + 2 - 0. 26. x - (2 - \/3)y -1 + vl -0,«-(2 + V3)v - 1 -VI - 0. 27. 3s - 4y + 24 - 0. 28. 3s + 4y - 24 - 0. 29. x - y - 5 - 0, a? + y - 13 - 0. 80. -2.0225. 31. 3. 32. x - 2y + 7 - 0, x + 3y - 8 » 0. 83. s-(2+\/3)y+ll+6\/3«0, s-(2-\/3)y-4- v"3 - 0. 34. x + 3y - 15 - 0, Zx - y - 5 - 0. 88. *\/5, (1, V). 36. (3, 4). 37. 2x - y + 4 - 0. 88. 3s - 4y - 0. 89. 3s + Zy - 13 - 0, 3s + Zy - 11 - 0. 40. 3s - 4y + 1 - 0, 3s + 4y - 7 - 0. 42. 4s - Zy + 6 - 0, 3s - 4y + 15 - 0. 44. s- V3y - 0, s + \/3y = 0. 45. s - y + 1 - 0. 46. s - VZy - 0. 47. y = 4, y - 3, s = 2, s « 3. 48. (^, ^} . 49. s + (8 - 5 \/3)y - 50 + ZOVz - 0. 50. (a -f b)x — (a — 6)y — be — ac - 0. 51. 3s 7 + 5y' + 4 = 0. 52. 21s + 77y - 1 - 0, 99s - 27y - 79 - 0. 54. (1) A - £ or C =* 0. 55. (1) 116° 84\ (2) A » -£. (2) 79° 42'. (3) A+ 2B + C - 0. (3) 60 d . 56. 8 + 6\/3. 57. (-1, 1). 58. 3. 59. 8s - by + 30 - 0. 60. 175° 26'. 61. 18s + 129y - 50 - 0, 138s + 79y - 210 - 0. 62. s + Zy - 30 = 0, s + Zy + 10 - 0. 63. y + 6 - 0. 64. (2, 4). 65. s + y - 2 = 0, s - (2 + V3)y - 2 - 2\/5 - 0, s - (2 - V3)y - 2 + 2V§ * 0. Page 88. Art 74. 1. 5. 7. 9. (l, 2); (i 1); <-*, ■ (-3a, 3. t. (- iV2- - 1); o. -ia);iV7, 2, 3); 1. i. 8. (-6, -J e. (-i, 8. (a, 3a) 10. (Ja, - J); 2. 4. -i);iV2. ; 3a. |a);Ja. 0, 2); i Pages 90-92. Art. 75. 1. s* + y* - 6s - &y + 20 = 0. 2. s* + y* - 2s + 6y + 5 - 0. 3. s* + j/» + 4s - 4y + 3 - 0. 4. s» + y* - 2s - 2y - 23 - 0, Digitized by Google ANSWERS 821 5. a;* + y* + 2x + 2y - 23 - 0. 6. a* + y* - 4a: - 6y - 12 - 0. 7. a;* + y* - 2y - 12 - 0. 8. a;* + y* - 2x - 24 « 0. 9. ** + y* - 4a? - 8y + 10 - 0. 10. a;* + y* + 4a; - Sy + 10 - 0. 11. a* + y* ± 8a; + 6y - 0. 12. a;* + y* - 24a; ± lOy - 0. 13. a* + y* - 20. 14. ** + y* - 2» - 4y - 8 - 0. 15. x* + y* + 2a; - 6y = 0. 16. a;* + y* - 4a; - 6 - 0. 17. x* + y* - 4a; - 4y - 17 - 0, a;* + y* - 10a; - 22y + 121 - 0. 18. a;* + y* + 6a; - 2y - 15 - 0, a;* + y* - 10a; - 14y + 49-0. 19. x* + y* - 2a; - Sy - 3 » 0. 20. a;* + y* + 4a; - 6y - 0. 21. a;* -f y* - 2a; - 2y - 3 = 0. 22. a:* + y* - 6a; ± Sy + 9 - 0. 28. a;* + y* ± 4a; - 8y + 16 - 0. 24. x* + y* - 4a; - Sy + 10 - 0. 25. a;* -f y* - 6a; - 6y + 9 - 0. 26. a?* + y* - 2a; + 2y + 1 - 0. 27. a;* + y* + 4a; - Ay + 4 - 0. 28. a;* + y* + 6a; - 6y + 9 - 0. 29. a;* + y* - 4a; - 4y - 2 - 0. 80. ** + y* + 4c - 6y + 8 - 0. 31. 4a;* + 4y* + 20a; - 20y + 25 - 0, x 1 + y* + 30a; - 30y + 225 - 0. 82. x * + y* - 10a; - lOy + 25-0. 83. a;* + y* - 10a; - lOy + 25 - 0, x* + y» - 26a; - 26y + 169 - 0. 34. 2a; - y - 7 - 0. 85. * - 2y + 5 - 0; 2V& 86. x - 3y + 5 - 0; 2\/l0. ' 87. x* + y* - 85 - 0. 88. 36a;* + 36y* + 84a; - 12y - 575 - 0. 89. a;* + y* - 65. Page 94. Art. 76. 1. x + 2y - 5 - 0. 2. x + y - 1 = 0. 3. x + ?y + 4 - 0. 4. 5a; + 3y - 7 - 0. 5. 5a; + 15y - 34 = 0. 6. 7x +9y -18 » 0. 7. x* + y* - 3x - 4y + 5 = 0. 8. x* + y* - 5x - 13y + 42-0. Pages 95, 96. Art. 77. 1. Circle, center at origin, r — 3. 2. Circle, center (J, |), r - JvTo. 3. Circle, center (V, 0), r = f. 4. Circle, center (^, - 1), r = ^VOS. 5. Two circles, centers ( ± }, 2), r - }. 6. •¥• 7. Circle, center (0, 1), r = V29. 8. Two circles, centers (0, ±2\/3), r = 4. 9. Circle, center (-9, 0), r = 10. 10. Circle, center at origin, r — 3. 21 Digitized by Google 322 ANALYTIC GEOMETRY Page 97. Art. 78. 1 p — 2. 2. p ~ 10 cos 0. 3. p » —8 cos 0. 4. p = 6 sin 0. 5. p - — 4 sin 0. 6. p - ± 12 sin 6. 7. p — ± 12 cos 0. 8. p - 6 cos (0 — i) • 9. p - VS. 10. p - 3 sin 0. 11. p « — f cos 0. . 12. p = 6 cos + 8 sin j0. 13. s* + y* + 6y = 0; (0, -3); 3. 14. s» + y» -4s = 0; (2, 0); 2. 15. s* + y* - s - y - 0; (J, J); JV2. 16. s* + y* = 25; (0,0); 5. 17. s* + y* 4- 2x + 3y - 0; (-1, - 1); iVl3. 18. s*,+ y* + 3s + 4y - 6 - 0; (- f, -2); {. 19. s* + y*-9 =0; (0,0); 3. 20. s* + y*-4 = 0;(0, 0); 2. Page 103. Art 85. 2. (J,0),(-i,0),(0,i),(0, -*);2s-fl.= 0,2s -1=0, 2y + 1 - 0, 2y - 1 - 0; 2. 5. (1) y* = 12s; (2) s* - 24y; (3) y* - -16s; (4) s* - -8y. 6. (1) y* = 8s; (2) s* - y. 7. s* - 5000y. Pages 106, 106. Art 86. 1. (1) y* - 8s - 8y + 40 - 0; (2) y* + 8s - 6y - 7 - 0; (3) s* + 12s - 12y + 60 •- 0; (4) s* - 4s + 6y + 22 - 0. 2. (1) (5, 4) t s - 1 - 0; (2) (0, 3), s - 4 - 0; (3) (-6, 5), y + 1 - O; (4) (2, -4J), 2y + 3 = 0. 3. y* - 10s + 4y - 36 - 0, y* + 10s + 4y + 44 = 0. 4. s* - 6s + 8y + 25 - 0. 6. s'* + 4s' + Sy' + 32 - 0. 6. (1) y* - Sx - lOy + 57 - 0; (2) s* + 8s - 8y + 32 - 0; (3) y* + 8s - 4y + 36 = 0; (4) s* - 6s + 8y + 41 - 0. 7. (1) y* - 3s - 6y + 15 - 0; (2) 4y* + 25s - 16y - 59 - 0; (3) 2s* - 8s + 9y - 19 - 0; (4) 5s* - 30s - 16y + 13 = 0. Pages 108, 109. Art. 89. 1. (1) (3, 2), (4, 2), y - 2 = 0, s - 2 = 0; (2) (5, -4), (4}, -4), y + 4=0, 2s -11'- 0; (3) (-1}, 2), (-1*, 3i), 2s + 3 = 0, 4y - 3 = 0; (4) (6, -2), (6, -2|), s - 6 - 0, 8y + 13 - 0; (5) (-A, 2), (-If*, 2), y - 2 - 0, 6s - 43 - 0; (6) (|, «), (|, f|), 2s - 9 = 0, 40y - 239 - 0. 2. 5s* - 9s - 2y + 4 = 0, (*,-*), (,A -A),, f . Digitized by Google ANSWERS 323 3. y* - 12* - 8y + 28 - 0. 4. 3y» - 25s + 18y + 77 = 0. 5. (1) y" - 4s'; (2) s" - -16y'; (3) y" - -8s'; (4) s'« - Jy'. 6. (1) 4s - 3 - 0, s' + 1 - 0, (|,3), (1, 0), 4; (2) y - 5 = 0, y' - 4 = 0, (4, -3), (0, -4), 16; (3) s - 3 - 0, s' - 2 - 0, (-1, 2), (-2, 0), 8; (4) 42y - 11 = 0, 12y' + 7=0, (-*, V), (0, A), J. Pages 110, 111. Art. 90. 1. X 7 * - -2py', y'» = 2ps'. 2. (1) a" + 2s'y' + y" - 4\/2s' + 4VV - 0; (2) 3s'» - 2\/3s'y' +y" + (6 \/3 -4)s' - (4\/3 + 6)y' + 24 -0; (3) 5y'« - 2\/5s' - 2\/5y' - 10 - 0; (4) 13s'* - 6a/13s' - 14Vl3y' - 68 - 0. 3. 9s* - 24sy + 16y» - 116s - 162y + 321 = 0. 4. (1) y"» - 3\/2s"; (2) s"* = AV^y"; (3) s"» - - iV2y". 3 -*^ Pages 111, 112. Art 91. 1 ± cos Page 118. Art. 92. 2. (3.31, ±7.28). Page 114. Art. 93. 1. s* = - 144 y, 24.31 ft., 22.22 ft., 13.89 ft. 2. 14' J", 11' 3", 6' 6}". 3. s» - a*f*y. 4. s» - H 8Jl (y - 20). Page 116. Art. 94. A 3t>* sin* a 5. (1) 23.67 mi.; (2) 20.50 mi.; (3) 20.50 mi. 7. 401.5 ft. per sec. Page 116, 116. General Exercises. 1. (621, 62500). 8. 1J in. from back of reflector. 4. (4, 2), (8, 8). 6. s - y - 2 - 0. 6. - 6}. 7. (3, 1), (42, 14). 8. 3jt_\/57. 9 . (ff _2i). 10. (1) y» + 4s - 4 = 0; 4 (2) 25y 2 - 60s - 36 - 0; (3) y* + 24s - 144 = 0. 11. (1) (12, 70° 32'), (12, 289° 28'); (2) (4, 90°), (4, 270°). 13. 2s* + 2y» - 5ps = 0. 14. 4\/3p. Digitized by Google 324 ANALYTIC GEOMETRY ' Pages 122, 123. Art 9t. 1. (1) 5,4,1, (±3,0), 3s ±25-0; (2) 10,6,1, (0,±8),2y ±25-0; (3) 3, 2, iV5, (±V5, 0), 5s ± 9V5 - 0; (4) 4, 3, i\/7, ( ± V7 t 0), 7s ± 16^/7 - 0; (5) 2V% y/l, Wl, (± V3, 0), 3s ± SVS - 0. (6) 3, V6, iVZ t (±V3, 0), x ± 3V3 - 0. 2. a. 3. (1) 4s* + 9y* - 144; (2) s* + 4y* - 16; (3) 5** + 9y* - 81; (4) 20s* + 36y* - 1125; (5) 112** + 256y* - 3067; (6) s* + 4y* - 64; (7) 2s* + 3y* - 18. 4. ± $ V21, ± 2f , 0, ± 1 V^ll. 6. 2s* + 3y* - 6. 8. (1; \/6, VB, i V6,|\/6;(2) V2^ VJf i y/% V2^~; (3) V«, Jv^i, - q Vrif=D, - q Vi; (4) VJ, Vp, J Vi?Frt. f V* 9. 3.8, 6.2. 10. 3s* + 4y* - 576. 12. (0, 0), s - ± «, 0. 13. (1, ± 2). 14. 6*s* + c*y* - a*6*. 15. (a* - 6»)s* + aV - a*(a> - 6*). Page 124. Art. 100. 1. (1) 9s* + 25y* - 54s - 200y + 256 - 0. (2) 48s* + y* + 288s + 14y + 445 = 0. 2. 4s* + 9y* - 40s + 72y + 100 = 0, 9s* + 4y* - 90s + 32y + 145 - 0. 3. (1) (-1, 4), (7, 4), 4s + 13 - 0, 4s - 37 - 0;' (2) (-3, -7 ± J VT41), y - - 7 ± H VUT. 4. 7s* + 16y* - 140s - 64y + 512 - 0. 5. 16s* + 25y* - 96s - 200y + 144-0. 6. 2s* + y* - 16s - 4y - 0. 1. (1) (s + 1)» Page 127. Art. 101. -2)* -l; (-1, 2); (^4,2), (2, 2); (-5, 2), 16 r 7 (3, 2); a = 4, 6 = \/7; 3s + 19 - 0, 3s - 13 - 0. (2) to^W+fX^Lh'-i. (4, 1); (4, 1 ±2^2); (4, -3), (4, 5); a - 4, 6 = 2\/2; y - 1 ± 4\/2 = 0. (3) ^=-^ f + &+J) 1 - l; (1 - 1); (1 ± i V5, -1); (J, -1), (!, -i>; o = i, 6 = J; s - 1 ± a V5 = 0. Digitized by Google ANSWERS 325 (4) £+»'+ (JLZ.3)' , 1; (-!, 3); (-i ± J VB, 3); (-1 ±tV5, 3); a « $\/5, & - VI6; * + 1 ± *Vs - 0. 2. 6s'* + 7y'* = 8. 3. (1) *J + *P - l f (±3, 0), 3s> ± 16 « 0. (2) § + ^ - 1, (0, ±2 \/2), y' ± 4 V2 = 0. (3) j + j - 1, (± 1 \/5, 0), 10s* ± 3 V5 - 0. ' W ^ , + £ " *' (± * ^ 0) ' 2s' ± 9 \/5 « 0. 4. 9s* + 25y* + $4s - 200y - 873 - 0; 5. 3s* + 4y* - 24s - 16y + 16-0. 6. 3^* ■+ 4y'* = 36. 7. 3s* + 4y* - 20s + 12-0. Pages 128, 129. Art. 102. 1. (1) (a* + 6*)s'* + 2(a* - 6*)sV + («*+ W* - 2a*6* - 0, (2) 43s'* - 14\/3sy + 57y'* - 576 - 0; (3) s'* + 9y'« - 36 - 0; (4) 9s'« + 2 V3sy + 1 ly'* + (6 - 8 V3)s' + (6 Vs + 8)y' - 40 - 0; (5) 3s'« + y'* + 3-v/2s' - VV +4=0; (6) 38s'* + 12sV + 22y'« + 2\/5s' - 21 VV = 0. 2. ^ + | " 1. 3- 9*"* + 3y"» -82-0. Page 129. Art. 108. 1 a . e P n . gP- 4 o* ^ fl * (l - e> > * 1 - e ops ^ ' 1 + e cos * * 1 - e 2 cos* Pages 181, 182. Art. 106. 1- ($)*+& = 1, (±m.8,0). 2. ^ + ^ = 1,(190.75,0). 8. 36.37 ft., 27.78 ft. 8. 0.95+ ft., 4.02- ft., 10.16- ft. 6. 13,000 mi. 8. 45.1 in. per see. 0.14 in. per see. . Pages 132, 183. General Exercises. ** 36 + 16 lm 2 * 72 + 144 " lm 5 * 72 + 36 " 1b Digitized by Google 326 ANALYTIC GEOMETRY I 6, 2\/6; (±2a/3, 0); i\/3; x ± 6 a/3 =0. ?. 5, 4; (-1, -2), (5, -2); {; 3x + 19 = 0, 3x - 31 - 0. •. 9*'» + 4y" - 36. 9. 25x" + 16y" - 400. x* , y* 9 11. 189** + 96xy + 161y» - 1494x - 25Sy + 2106 - 0. 12. 6.4. 10. i + ~r— n - 1. a > 3 Pages 138, 139. Art. 110. 1. (1) 10, 8, (±V41, 0), 41x ± 25V41 - 0; (2) 12, 20, (±2^34, 0), 17x ± 9\/34 = 0; (3) 6, 8, (±5, 0), 5x ±19 - 0; (4) 16, 12, (0, ±10), 5y ± 32j== 0; (5) 2V2, 2\/3, (0, ±\Z5), 5y ± 2\/5 - 0; (6) 10, 6, (0, ±\/84), My ± 25V34 - 0. 3. 0, ± 5J, ± i V^5. *- 5 i~ 5. 5x* - 9y« - 36. 6. 5x« - 4y» - 80. 8. (1) 4, 3, J, f; (2) V6, 2y/% iV21, f V6; (3) 1, 4, Vl7, 32; (4) V2m, Vmj JV6, V2mj (5) Vq f Vp, ^ Vpq + q*, ~ VgJ 9. (1) 3, 4; (± 5, 0); #; 5x ± 9 - 0. (2) 6, 2V6; (± 2\/l5, 0); (i Vl5; 5x ± 6\/l5 - 0. 10. x* - 3y 8 + 3 - 0. 13. xV = 8. 14. 7x» - 9y» = 1008. 15. x» - 3y» - 144. Pages 142, 143. Art. 113. 1. (1) 2x ±V6y - 0; (2) x ±V2y = 0; (3) x ±V2y = 0; (4) 5x ± 4y - 0; (5) x ± y - 0; (6) x ± y - 0. 2. 3x« - 4y« + 48 - 0, (± 2\/7, 0), (0, ±2\/7),' 7x ± 8\/7 - 0, 7y±6V7=0. 5. 2x'y' - a*. 6. x* - 3j/« - 16. 11. ±4\/2. 12. ±0.9014. Digitized by Google ANSWERS 327 Pages 148, 144. Art 114. 1. (1) 9x* - 2by* - 72x - 150y - 306 - 0; (2) x« - 4y« + 12* - 16y + 36 - 0. 2. (1) (-1, -3), (9, -3); (4 ± V34, -3); 34x - 136 ± 25V34 - 0. (2) (-6, -4), (-6, 0); (-6, -2 ± 2Vb); by + 10 ± 2V5 - 0. 3. 64s* - 36y* + 256x + b04y - 1283 - 0. 4. (1) 9x» - 2by* - 72x - 150y + 144 - 0; I (2) x* - 4y» + 12x - lQy + 4 - 0. 5. (1) 3x - by - 27 - 0, 3x + by + 3 - 0; (2) x - 2y + 2 = 0, x + 2y + 10 - 0. Page 146. Art. 115. 1. (1) ^jff^* " -^T^' - 1; ». 3); (1, 3), (11, 3); (2, 3), (10,3); a - 4, 6 = 3; 5x - 30 ± 16 - 0; 3x - 4y - 6 = 0, 3x + 4y - 30 = 0. (2 ) fiLzJ) 1 . &L+i> f . i ; (-3, f); (-3, | ± Vl7); (-8,1), (-3, J); a - 1, 6 = 4; 34y - 85 ± 2\/i7 - 0; x - 4y + 13 ^ 0, x + 4y - 7 = 0. (3) &L=« f . &L+l) f = i ; (j - } ); ( _ f| _ j), (i, -i); (i ± \/7, -J); a - V7, & = V2; 6x - 17 - 0, 6x + 11 = 0; 14y + 7= ±2Vl4xTVl4. (4)^^) f « &L=Li> f . 1; (1| 3); (1, 3 ± \/l7); (1, 3 ± 2V2); a - 2\/2, 6 = 3; 17y - 51 ± 8\/l7 - 0; 4x - S\/2y - 4 + 9\/2 = 0, 4x + 3\/2y - 4 - 9 V2 - O.n (5) ^S^'"^^^ 1 ^" 2 ' 4 ^"^^ a - 2\/7, b - V21; y - 0, y - 8 - 0; 2x - V3y + 4 + 4\/3 - 0, 2x + V3y + 4 - 4\/3 - 0. 2. 9x" - 25i/'« 4- 225 =0. 8. (1) ^- ^* - 1, (±5, 0), 5x' ±16 = 0; (2) *£- ~ r 1, (0, ± Vl7), 17y' ±Vl7 = 0; (3) y - ^ - 1, (±3, 0), 3s'±7 - 0; Digitized by Google 328 ANALYTIC GEOMETRY (4) Y""? - *> (°» ±Vl7), MY ±SVT7 - 0. 4. «x» - 16y» + 54* + 128y + 1601 - 0. 5. 16x* - 25y» + 64* + 200y - 736 = 0. 6. 3s» - y* - 84x + Ay + 536 - 0. Pages 14ft, 147. Art 11C 1. (1) *Y - 8; (2) £*- b* *' (2) o» (3) x" ■ - y" - 16; (4) n*'» + sovlteV - ■39y" - 676; »?• 3 *' (6) x>* - 3y" - 2a/ + 6/ - 11 -0. (Dv- i; (2) *>« 30« *?-£'- ™' 3 4 1. 3. *"« - y" 1 - ll\/2. 4. (-3, -1); (2.15, 1.13), (-8.15, -3.13); a? - (1 + y/2)y + 2 - V2 - 0, x + (\/2 - l)y + 2 + \/2 - 0; x - (\/2 - l)y + 4 - y/2 « 0, x + (V2 + l)y + 4 + V2 - 0. Paget 147, 148. Art 117. *• * 1 + e cos p 1 T e cos * Paget 149, 150. Art 118. 4. 5x* - 4y» - 20. Pages 151, 152. Art 119. 1. pv — 10. 2. t#Z = 20. 8. One branch of an hyperbola with foci at centers of circles and transverse axis equal to the difference of radii. 4. Hyperbola with foci at centers of circles and transverse axis equal to the sum of radii. 5. An equilateral hyperbola with the ends of the base as vertices. 6. 345 ft. at an angle of 11° 43' with the perpendicular to AB. Digitized by Google ANSWERS 329 Paget lfS, lit. General Exercises. 1. a - \/6, b - 2; WlS; (±\/IO, 0); 3y* - 2x* - 12. 2. (±2\/3,2),(±2\/3, -2). _ *. a - 6, b - 4; (3, 2 ± 2\/l3); J\/l3; 13y - 26 ± 18\/i3 - 0. 7. (y - 2)» - 0. 8. («V39, fjVS), (-»*V39, -«V39). 9. 7.806+, 17.806+. 10. s» - 3y* - x + Zy - 0. 11. 4y» - x» - 11. Page 198. Art. 122. 1. Hyperbola, 3s"» - 2y"» - 6. 2. Ellipse, 9s"* + 16^ - 144. 3. Ellipse, 3s"» + y"» + 6-0. 4. Parabola, 2y"» - 3a;' 4 . 5. EUipse, s" + 4y>« - 16. 6. Parabola, y'* - 3s'. 7. Imaginary ellipse, 121*'" + lly"» + 199-0. 8. Two lines, 2&r"» - y"* - 0. 9. Parabola, *"* - fVoy". 10. Parabola, y"» - H Vl3s". Page 160. Ait. 124. 9. Jfe - 0.0000082, p - 0.G000082P- 4 ". 10. c « 6028, *»*•* - 6028. Page 162, 163. Art. 126. 17. (1) x» + y» - 2aj4nj£s + « f - °J ( 2 )*L + y% =1- (3) p' ■ TT¥> a; 1 Jfci" *2a» y - y 1 20. J(4a* - Jb«) cos 20. 64 2304* Pages 173, 174. Art 136. ID. 2r. 11. No. 12. y = 18 sin fcrf. 13. y - 8sin(M20° + 65°), 3, J. 14. (1) 3.34, 0.299; (2) 16, A. 16. fcr f fcror 1 sec., 2 sec. 17. J*, 0. 20. 0.0007854. Pages 179, 180. Art. 140. 27. p - r cos $. 28. p = 2(6 + r cos 6). Digitized by Google 330 ANALYTIC GEOMETRY Page 182. Art 14L l.x«-3 + *,y-2 + 2/. 2. x - 2 - 5 c<* #, y - 3 - 5 sin 0. 4. y* + x + 2y-3 -0. §.^ + ^-1. 6. 4x» + y» - lfix + 12 » 0. 7. x* + 2xy + y* - 2x + 2y = 0. 8. te* + 4y«-90x-32y + 253-0. *. y« - tx. 10. x* - J(y + 1). 1L x» + y* - a» + &». 12. (?) * + &\ * - 1. Page 183. Art. 142. 3. x — a$ — 6 sin 0, y = a — 6 cos 0. Paget 193-195. Art. 161. 1. I = 0.00102T. 2. y - 1.405x + 7.527. 3. y - 1.403x + 7.54. 4. R = 0.00313/ + 9.8753. 5. H - 03119/ + 606.00. 6. # = 0.1470TF + 1.7957. 8. TF - 0.5015/ + 54.10. Paget 201-203. Art. 166. 2. y - 0.9975X" 1 " or, very nearly, y = x -1 - 17 . 3. ff - 3.867Z>>-™ 659. 4. p = 30e-o.ooooi* # 6. m - 0.1374p-° M7 . 8. m - 0.00014V°-". 9. pv l " - 147. Pages 204, 205. Art. 167. 2. y - 0.5 + 0.02x + 2.5x* - 0.3x». 3. / = 132 + 0.875x + 0.01125x». Page 210. Art. 160. 1. x + 2y - 1 = 0. 2. 2x - 6y - 3 = 0. 3. 3x + 4y + 1 - 0. 4. 2x + 12y + 5 - 0. 5. 3x - 2y + 6 - 0. 6. 3x + 2y - 4 - 0. 7. 3x + 4 - 0. 8. x + 14y + 17-0. 9. (1, 3). 10. (5, 2). 11. (1, 2). 12. (4, 2). 13. (3, 1). 14. (3, 6). 15. (1, 1). 16. (-1, 3). 17. (-2, -8). Pages 214, 215. Art. 166. 1. x + 4y - 0, 3x - 2y - 0. 2. x + y - 0, x - 6y - 0. 3. 3y - 2 - 0, y + 4 - 0. 4. 2x + 9y - 20 - 0. Digitized by VjOOQIC ANSWERS 331 5. 2x-9y + 16 -0. 6. (|, -*),(-!, |). 1 o.(^^),(-^-^).i a .,-« y + 2 -o. 13. 3x - 2y - 12 « 0. 14. 2x - 3y + 16 - 0. 18. Ellipse concentric with original ellipse, major axis « V^2o, minor axis = V2&. P««es 218-220. Art 187. 2. 4.5, 0.0802. 3. A = x», 13. 4. A - tx\ 10.25r. 6. A = Jd* d - V2A. 6. C - 2\/iSf, S - ^36>F*. 7. 7 - xr*A, h - -^ :, S - ^f. 8. -4, 44, -12. 9. 1, VlO, 7. 10. 2, 1.6778, -1.9208. 11. iV3, 0, -J. 12. Sx*y - 4xy» - 2y*, 4zy* - 3x*y - 2y\ -2y* - 3x*y - 4xy*. 16. a; - sin-*, y, x = ^-|. 16. a; - ± jV& 2 - y\ y - ± ^Va s - a* 17. x = (a* - y*)», y - (a* - x*)». 18. x = (a 1 - y 1 )*, y = (a 1 - a; 1 ) 1 . „. , x — 2o / — 19. y = ± — - — V ox. 20. a? - ± lV-2y*±Wy* + 256, y - ± ?\/4 - a:*. 21. a? - 1 ± iV3 + 2y - y*, y - 1 ± 2\/2x - x*. 22. a; - ±*-i^\/l6 - y\ 23. i - *-*' cos 2*. 24. *> = ± 7= ,0 - i cos"" 1 — 2 - Vcos 20 «> 2 25 . ^q cos ^ _, ( r 2a ± V4a« + <p* = cos" 1 y- sin* \ <p / 26. <p - ± aVtan 0(3 - 4 sin* 0). 27. x = ^> y - a»x. 28. x - jV2(y + Vl - y»), 2/ - iV2(* - Vl - **). Page 226. Art. 171. 2. 4, -i. 3. 2, 2x - y + 1 - 0, x + 2y - 7 - 0. 4.6*. 18*'. 6.2. l.~±. S.f. t-nj^T? Digitized by Google 332 ANALYTIC GEOMETRY ID. 3xj* + 4*i. 1L ftri* - 8*1 + 6. 12.-^. IS. =*. 14.-. 16. 2yi 9yi y x - fo j 1)t 16 - 2» - 2y + p - 0, 2s + 2y + p - 0. 17. -0.4364, 0.4364. 18. 73° 44.4'. 19. 120°. Paget 131-233. Art. 181. 1.^-te. dx 8. ^ - 20s«. dx a**- 28 ** . <fy 2a ax 3** 5 <fr ,. 9 * <te 16*** *. * « 2 , * dx 3 7 ^?- > dx x* ax 5x** ft * - 2 10.*. • dx 3** dx 2 "••s-* dt fit* 16. * - }t*. 16. J - 4x« + «x. if.J-a.-a. 18. ^ - 3x« - Jx* + 3. 19 d„.3x. + l. dx 2x* »*-*+Jr 21. J - 6(2x + 1)«. 22. * - 24*<3*» + 2)« - 2. as ^ _ * -a ^ dy m 4x - 7^ dx V2x + 3 dx 2 V2x» - 7* M dy = 2x + 7 »-t --!-.• dx 3^0e* + 7x - 2)» 27 ^ - - 1. " dx x 4 dx x* 29 * - 2 • ,0 * . 1 . * ¥ * dx (x + 1)* dx 2\/(x + 1)» tt4 dy 20x 31 " dx " (x - 1)»* 22 *«= 2 M " dx (x + 1)« Digitized by VjOOQK ANSWERS 333 »*..'■ k.*- 1 — *• dx (x - 3)»* •** dx (*• + 1)«* 36. P - 21x« - 24s* + 12a* 36. ^ * dx <fc 2\/x + 1 2\/x - 1 87 *U » 0s» + 14x - 3 # dx * 2 V3x» + 7s» - 3x + 2 33 dy « 2ox + b _ 1 " dx " 2\/ax l + bx + 6 2\/* + d ^ dy 4x 4 + 10x ^ dy 2x TO - dx" -^JT+5 • ^ * " <* - 1)» 41 *. \ 4f. *t . - «' ' * <& ( x + l)Vx« - 1* dx V(x» - a»)« 43. *■■ ^= + -37=1=. 44. *- -L-i + l*.. 45. jjj-»5x< -3x* + 2x-2. 46. ^- (x + a)« fc >(* - 6) w ->(w« + nx + dm - 6n). 4t. ^- (x + l)*(2x - 1)*(16* + 1). 43. dy _ 2-4* 4ft d« n*»-> 50 dx (x - 1)** w# dt (1 + 0» +1# dt 2« dt «* + 1) VF^l 51. 0,3, 12. 63. 1:4, 1:8, 1:16. 54. x - y - 6 *= 0, * + y + 6 « 0; 29x - y - 38 - 0, x + 29y - 582 - 0. 55. At the points whose abscissas are r * o M Xi* — 1, x Xi 1 , . 66. y - yi - ^ (x - x,), y - y, - j _ — (a; - Xi). 57. (-1, -6), 7x + y + 13 =0. 58. At (1, 0) at 135 % , at (-3, -4) at 18* 26'. 60. 1.0026026. Page 234. Art. 182. 1 *£ - 5? q dy 3x» 4- 1 x " dx * y«* *' dx 3y* + l" # dy 6%c . dp p dv v dx a*y dv v dp p Digitized by Google 334 ANALYTIC GEOMETRY k *M _ 4x» - 8xy», dx _ da? 8x*y — 3y* ' dy ™ dy jr| dx x^ dx* x i' dy~ ""y** dy x +Va? 1 - y» e, i-- 8x«y -3y» 4x* - -8xy* 7 * T# dx *♦ "F > *«(t> - ■W y dp gy — 2ab — pp* d» _ ______ . dti m v»(v — 6) ' dp = o» — 2a& — p** # 10. 3x + 4y - 26 - 0, Ax - 3y = 0. 11. x - 6y + 17 - 0, Ox + y - 9 - 0. 12. 8x + 5\/6y - 36 = 0, 25x - sV&y - 18 - a 19. xVyi + y Vxi - Vasiyi - 0, xVii - y Vyi - XiVxi + 2/iVyi - 0. Page 237. Art. 185. L Rising for all values of x. 2. Rising for x > — 2, falling for x < — 2, 3. Rising for x>0, never falling. 4. Rising for x>0, falling for x<0. 5. Rising for all values of x, except x = 0. 6. Falling for all values of x except x = 0. 7. Rising for x > V3, and x < — V3, falling for - V3 < x < V3. 8. Rising for x > 1+ 8 and * < "1 V , falling for 1 ~ 3 < a? < 1 + 8 • •• Rising for x >1 and x <J, falling for } < x < 1. 10. Rising for — 1 < x < 1, falling for x > 1 and x < — 1. 11. Rising for — 1 < x < 1, falling f or x > 1 and x < — 1. 12. Rising for x>2 and x< — 1, falling for — Kx<2. 13. Rising for x>— 2, falling for x < — 2. 14. Rising for x > 1 and — Kx <0, falling for x < — 1 and <x <1. 15. 278, 19, 3, y decreasing twice as rapidly as x is increasing. 2 ± V'22 16. 5 no real values of x. Page 239. Art. 186. 1. Min. at x = 0. 2. Min. at x = 2. 3. Max. at x - 3. 4. Max- at (0, 2), Min. at (0, -2). 5. Max. at x = }(1 - Vl3), Min. at x - |(1 +Vl3). 6. Max. at x - 0, Min. at x - 4}. 7. Min. at x - 16. 8. Min. at x - 3. 9. (4.5, 4.1). 10. v% * m * " . Pages 241, 242. Art. 187. 1. Upward x>0, downward x<0, Infl. at x « 0. 2. Upward for all values. 3. Upward x>0, downward x<0, Infl. at x =* 0. 4. Upward Digitized by Google ANSWERS 335 x <0, downward x >0, Infl. at x = 0. 5. Upward x > 1 and x < — 1, down- ward — Kx<l, Infl. at x = ±1. 6. Upward x>l, downward x<l, Infl. at x » 1. 7. Upward a?> J and x<0, downward 0<x<}, Infl. at x =» and x = |. 8. Upward x>$, downward x<4, Infl. at x = J. 9. Upward x>J\/3 and x<-J\/3, downward -i\/3 <x < iVz, Infl. at x - ±i V3. 10. Upward x> 4 + 3 V ^ and x < 4 " j^*' downward 4 "/ 1 "^ x < 4 + /", Infl. at x = 4 * ^. 11. 3 3 3 -2, Max. at x = 1 - J\/6, Min. at x = 1 + }\/6. 12. 0, 2, 20 in. per sec. 13. No points, 1, 2. (6s + 2)dx. 9xdx W Pages 246, 246. Art. 190. 5. -0.15,0.25. 6. -0.09428, 0.09428. 7. 8. (3x» + 4)dx. 9. (4x» - 9s* + 4x)ox. 10. 11 *** . 12. -^dx. 18 Vx* + 4 xt u *** . 15. 0*o7, 64. V(x* + 5)« Pages 249, 260. Art. 198. 11. x* = iy. 12. x' - Zy -f 18. 2x ! - 3y + 4(3 - V2) « 0. 14. Af V2 square units. 15. V\/6 square units. 17. 64 square units. 18. 20} square units. 19. 25$ square units. 20. 3V2 square units. 21. J square units. 22. 2} square units. Pages 253, 254. Art. 197. 1. / ^ 3 cos 3x. 2. / - sin 2x. ax ax 8. ^ - -2 sin (2x + 1). 4. ^ = cos 2x. 5. ~ = 3 cos 3x cos 2x — 2 sin 3x sin 2x. ax 6. P = 3 sec' 3x. 7, ^ = sec* x. ax ax 8. ,- = 15 tan* 5x sec 2 5x. . 9. ~ = x cos x + sin x. ax ax Digitized by Google 336 ANALYTIC GEOMETRY 10. ^ - (3x* + 2x) cos (x« + X*). • 11. ^ - (2x + 3) cos (x« + 3a: - 4). 15. g| - -3 sin (fix + 4). IS. ^ - } cot 3xV«n3x. 14. ^ - } sin 4a:. 15. g - J tan xVsec x(3 cos* x + 1). 16. -g — — mng(cot* +1 gx + cot*"" 1 qx). 17 dy = 2 sin a?. * da: (1 + cos x)* d0 1 — sin 30 d$ 20. & - 3 cos x(l - 4 sin* x). 21. ^ - cot i*. ax ax 22. 2 square units. 23. 0.7071, -0.4161. 24. 1,0, oo. 25. Max. at x « (4n + 1) =» Min. at x «■ (4x + 3) ~» Int. at x » nr. 26. -i cos 3x + C. 2(7. -f cos (3x - 1) + C. 28. i sin 4x + C. 29. i sin (4x - 2) + C. 30. J sin* x + C. 31. i sin 4 x + C. 32. -i cos* x + C. 33. — r^r sin* +1 x + C. 34. y = sin x. n + 1 Paget 259, 260. Art. 203. dy 2x-H7 . (2x + 7)dx *• dx " x* + 7x' y * x* + 7x 2 dy _ 0.8686 ? rf _ 0.8686dx dx ™ x ' x a dy = _1 , ^ _dx dx ~~ x x 4 dy = 0.8686 t rf _ 0.8686dx dx x x 6. ^ - 2xe* f , dy - 2xe**dx. 7. ^ - 6xe** ,+ <, dy - 6xe** ,+ «dx. Digitized by Google ANSWERS 337 8. 9. 10. 11. 12. 13. 14. 16. 16. 17. 18. 19. -X — e*(sin x + cos x), dy j2 - 2a* log. a, dy e*(sin x + cos x)dx. 2a te log« adx. 4.6052xl0* + «, dy - 4.6052xl0*+«dx. dy dx * ^ - 3x(3x - 2)'-> + (3x - 2)- log (3x - 2), <fy = [3x(3x - 2)'-i + (3x - 2)* log (3* - 2)]dx. *-*<*-«->,* |(e» - e-*)dx. d< — afce-«<, di — — a&e-«*d£ di RI -£! .. £/ _* 3i" ~T« L '*" "T 6 Ldt e""*(cos « — sin x)dx. 3^ = e~*(cos x — sin a;), < ax j - -J«-*«(6 sin 2* + cos 2/), dt _ -i<H«(6 sin 2* + cos 2t)dt. dy (x + 1)» (x + l)%fe dx = x* + 1 y x« + 1 dy dx ^ - (3 - 4* - tte*)«-» , l dy - (3 - 4x - 6x')e-*'dx. (4 + ?) (2x + log x), dy - (4 + D (2x + log x)dx. dy 2(x» + l) to+ * dx dy - 2(x* + l)** + » 2xM-_3x . x* + 1 2x» + 3x 21. 24. 26. 27. 29. 81. 88. 85. 87. 39. x* + 1 1, 7.39. 22. 0.4343, 0.0434 No Max. point, Min. (i, 1.193). 2.3026 square units. 6.693 square units. 3 log Cx. log C sin x, + log (x* + 1) + log (x* + 1) 23. dx. (1, 0.6931). 28. log C(x - 1). 30. x + log Cx. 32. ie** + C. ix* - 2x + log Cx. a 8 * 3ToJ^ + a y* - e» f . 22 84. »-ta«^ + i-Si- 36. x - 4<r* + C. 38. y = e»*. Digitized by Google 338 ANALYTIC GEOMETRY Paget 266, 266. Art 208. L 11. 2. 13. 8. V61, V85, Vl66. 9. (3, 4, 0). 10. (37, -36, 23). 1L 2: -1. la. (i, -l, -9). Page 269. Art 210 1. * VV* \/3, i \/3, or - iV3, - * \/5. - i V%' 1, 1, 1. 2. A Vio, - a Via - i VTo, or - ^ Vio, A Vio iVI6;7, -4, -5. 3.?,*, -*,or -$, ~M;2,3, -6. 4. -M, ~f,orf, -|,f. 5. A, - A, A, or - A, A, ~ A- e. A, A, -tt,or-A, -A,B. 7. «,A, -H,or-«, -*,«. 8. - 2, -9, 6. 9. (1, 0, 0), (0, 1, 0), (0, 0, 1). 10. (1, 0, 0,) (0, cos 0, cos y). 12. 45° or 135°. 18. 48° 51' or 131° 9'. 14. J\/2, i\/2, 0. 15. J V3, J, 0. Page 273. Art 213. 1. (2, 60°, 45?, 120°) or (-2, 120°, 135°, 60°). 2. (8, 60°, 120°, 45°) or (-8, 120°, 60°, 135°). 3. ( V3, 54° 44', 54° 44', 54° 44') or (- VS, 125° 16', 125° 16', 125° 16'). 4. 45° or 135°. 5. (8, 60°, 30°) or (-8, 240\ 150°). 6. (4, 210°, 120°) or (-4, 30°, 60°). 7. (4, 135°, 60°) or (-4, 315°, 120°). 8. 77° 56.6'. 9. f , - i, J, or -»,!,- f. 10. *. 11. -*?. 12. -f. Paget 277, 278. Art 218. U = 3. z « -4. 2. z «■ 0. y = 0. * « 0. 3. y s + z* - 9. x* + y* - 16. 4. 8s + 4y - 10* - 15 - 0. 5. y - z = 0. 6. 3* - y* = 0. 7. x* - y 8 - z* - 0. 8. s* - 4x + 8y - 6* + 29 - 0. 9. x* + y* + z* - 3* - 4* - 0. 10. x» + y* + s 8 - 6x - 72 = 0, x» + y* + ** - 2x - 80 = 0. 11. 17x« + 17y* + 17 2 « - 90s - 25Sy + 265 - 0. ** + y % + z* - 2x - 2y - 47 - 0. 12. x* + y* + « l - 2x + 2y - 4* - 75 = 0. 13. x* - y* - z» = 0. 14. s» - y» + «» - a 22 Digitized by Google ANSWERS 339 16. x* - y* - 2* = 0. 16. x - y* - z* - 0. 17. x* - 4y* - 42* - 0. 18. x* + 2/* - 2* = 0. 19. x 4 + y 4 + s 4 + 2x V + 2xV + 2y V - 4y* - 4** * 0. 20. s* + y* + 2* - 2z - 0. 21. x» + y* - sin 1 * = 0. 22. x* - (sin-ty)* + 2* - 0. »• *! + £ + £ - l- * 4 - -! + r! + -i * L a 1 6* 6* a* b* a* 25. (x* + y* + «* + 12)* - 64(y* + z*) - 0. Pages 281, 282. Art 221. 4. 3y* - 5** + 2 = 0, 3x« - 22» - 10 - 0, 5x* - 2y* - 18 - 0. 5. 3y - z % - 0, 3x* - 2* - 0, x* - y - 0. 6. z 4 + a*y* - oV = 0, 2* + ax - a* = 0, x t + y* - ox « 0. 7. y s - z* - 0, x' + 2* - a* - 0, x' + y' - a* =0. 8. m*y* + 2* - a*m* - 0, 2 - mx = 0, x s + y t - a* = 0. 9. 2 s - 3y* = 0, 2* - 3ox - 0, y s - ox - 0. Pages 291. Art. 229. 21. x s + y* 4- 22* - 4. 24. x* + 2» - 2px + p* - 0. 1. 8. Page 297, 298. Art. 287. 1 2 3*-3 y - - 3 2 3 -u, 4 + _ 2 -h_ 2 - 2 _. 1 S' + I^" , 2 „_* -n *-l^4. JL - 4 _L 7 9 X + 9 y - 4 ,1 A ^ I ^ 1* •9 2 + 3 =0 '^1 + ^ + i 12 1 ,12 18 „ x , y . T7*-T7 y + r7*-r7-°-i + -i8 + 1. 5. 2x - 3y + 22 - 1 - 0. 6. 3x + 2y - 2 - 4 = 0. 7. x - 2y ± 22 - 15 = 0. 8. 6x - 3y + 2 - 2 = 0. 9. x + 2y - 22 - 6 - 0, 91x - 122y + 462 - 318 - 0. 10. 3x - y - 4z - 1 = 0. 12. 3x - 4y + 22 - 4 - 0. 14. 3x - 4y + 42 - 16 - 0. 16. 3x - 4y + 22 + 29 = 0. 18. 56° 15' or 123° 45'. 20. 67° 7' or 112° 53'. 21. 5x - lly - 82 + 14 - 0, 23x + 25y - 2O2 - 28 - 0. 22. x - 2y - 22 + 5 = 0, 4x + y + 2 - 7 - 0. 23. x + 2y - 52 - 5 - 0, llx - 8y - 2 - 13 = 0. 24. ±6. 25. (2, -1, 2). 26. 31° 1', 64° 37', 73° 24'. 27. 16° 36', 25° 23', 58° 59'. 11. x + y + 2-3 = 0. 13. x-2y + 2-1=0. 15. 3y - 42 + 5 - 0. 17. I. 19. 48° 11' or 131° 49'. Digitized by Google 340 ANALYTIC GEOMETRY Paces MI-MS. Art Ml. 1. (1, 1, 0), (2, 0, -3), (0, 2, 3). 5. (2. -3, 0), (8, 0, -3), (0, -4, 1). t. (3, -2,0), (1,0, -1), (0, 1, -|). 4. (-1,1,0), (-1,0,3), (0,2, -2). «. (1, J, 0), (2, 0, 1), (0, 1, -1). 6. (1) * + y - 2 - 0, 3* + z - 3 - 0. (2) * - 2» - 8 - 0, * + 2* - 2 - 0. 7. (3) * + y - 1 - 0, y + 2z + 2 - 0. (4) 2* - - ' - - - - y + 2-0, 5y + 2« - 8 « 0. g — ■ ' 3 «-l "• W -1 1 3 9. 11* + 5y - 3 - 0, 3* - 5* + 11 - 0. 10. 2x + 5y - 1 - 0, 2x - 5s + 14 - 0. 11. * - 2 - 0, y + 2s + 5 - 0. 12. x — 2 « 0, y - 5 - 0. 15. x + 3y + 8 - 0, 2x - 3s + 10 = 0. 14. x - 3 - 0, y + 1 - 0. 15. * - 2 - 0, s + 3y - 0. 16. x + y - 2 - 0, 3s + z - 11 - 0. 17. 2x + y - 5 - 0, x - s - 1 - 0. 18. Zx - 2y - 0, 2* - s « 0. 19. \. 20. 0. 22. i, -}, }. 28. x - y + 2s - 7 - 0. 24. 7a? - 12y + s - 14 - 0. 25. x + y + s - 3 « 0. 26. 2s - y + * - 1 - 0. 27. 2s -3y- 4* + 6-0. 28. '* + * = ^- 2 = i + i Digitized by Google \ INDEX Numbers refer to pages. Abscissa, 13, 14 Algebra, formulas of, 2 Algebraic, equations, 154 hyperbolic type, 159, 197 parabolic type, 158, 195 Algebraic functions, 225 Amplitude, of function, 169 factor, 169 Analytic, geometry, 1 methods, 39 Anchor ring, 277 Angle of lag, 172 Angles, bisectors of, 70, 297 definition, 24 direction, 267 formed by lines, 24, 271 formed by planes, 295 tangent of, 27 vectorial, 30 Applications, of ellipse, 131 exponential functions, 164 hyperbola, 150 parabola, 113 straight line, 79 Arch, parabolic, 113 elliptic, 131 Areas, by integration, 247, 252, 254, 259 of polygon, 38 of triangle, 36 Asymptotes, 139 Axes, coordinate, 12, 261 polar, 30 rotation of, 34 translation of, 33 Axis, conjugate of hyperbola, 137 major, of ellipse, 120 minor, of ellipse, 120 of parabola, 102 of symmetry, 50 transverse, of hyperbola, 137 B Bisector, of angles formed by lines, 70 of angles formed by planes, 297 Boyle's law, 150 Cardioid, 185, 187 Cassinian oval, 163 Catenary, 167 Circle, equation, 86, 87, 96, 279 imaginary, 87 locus problems, 94 point or null, 87 radical axis, 94 satisfying three conditions, 88 systems, 92 Cissoid of Diodes, 160 Concavity, 236, 239 Conchoid of Nicomedes, 162 Condition second degree equation represents two straight lines, 77 Conditions for locus, 45 Cone, 290 Confocal ellipses, 132 341 Digitized by Google 342 INDEX Conic sections, 98 Conicoids, 284 Conies, definition of, 99 degenerate, 157 diameters of, 210, 214 directrix of, 99 eccentricity of, 99 focus of, 99 Conjugate, axis of hyperbola, 137 diameters, 212, 213 hyperbolas, 141 Continuous, functions, 235 Coordinates, axes of, 12 cartesian, 12 oblique cartesian, 15 origin of, 9 polar, 12, 29, 269 rectangular in plane, 13 rectangular in space, 261 relation between, 32 spherical, 270 transformation of, 33 Constants, 44 arbitrary, 53 Construction of, ellipse, 129 hyperbola, 148 parabola, 112 Curves, concavity of, 236, 239 cycle of, 169 empirical, 188 falling, 236 in space, 278 maximum point of, 235, 237 minimum point of, 235, 237 normal to, 223 periodic, 168 points of inflection, 236, 239 probability, 199 projection of, 280 proper sine, 168 properties of, 235 rising, 236 Curves, sine, 167 slope of, 222 tangent to, 222 Cycle of curve, 169 Cycloid, 182 Cylindrical surfaces, 274 Degenerate forms of conies, 157 Derivatives, 222 of algebraic functions, 226 of exponential functions, 254 of logarithmic functions, 254 of trigonometric functions, 250 Descartes, 1 Diameters, conjugate, 212, 213 length, 212, 213 of conic, 210 of ellipse, 210 of hyperbola, 213 of parabola, 213 Differential triangle, 245 Differentials, 242 definition of, 243 Differentiation, 225 fundamental method, 225 of implicit functions, 233 Direction angles, 267 cosines, 267 Directrices of, conic, 99 ellipse, 117, 120 hyperbola, 134, 137 parabola, 100 Discriminant, 2 Discussion of equations, 48, 174, 282 Distance, between two points, 17, 19,263 from point to line, 68 from point to plane, 296 Digitized by Google INDEX 343 Division, external, 19 internal, 19 of line segment, 19, 23, 264 Eccentricity of, conic, 99 ellipse, 100, 118 hyperbola, 100, 136 parabola, 100 Ellipse, applications, 131 center, 120 conjugate diameters, 212 definition, 100, 117 diameters, 210 directrix, 117, 120 eccentricity, 100, 118 equation, 118, 121, 124, 129 equation of tangent to, 234 focus, 117, 120 general equation, 128, 155 imaginary, 123 latus rectum, 120 major axis, 120 minor axis, 120 point, 123 sum of focal distances con- stant, 130 vertices, 120 Ellipsoid, 284 Elliptic paraboloid, 288 Empirical curves, 188 Epicycloid, 185 Equations, algebraic, 154 discussion of, 48, 174, 282 exponential, 163 exponential type, 197 general, of second degree, 87, 106, 124, 144, 154 graph of, 46 hyperbolic type, 159, 197 linear, 65, 19) Equations, locus of, 45 logarithmic, 165 of circle, 86, 87. 96 of curves in space. 278 of ellipse, 118, 121, 124, 129 of hyperbola, 135, 138, 142, 143, 144, 147 of line, 59, 62, 63, 64, 65, 78 of parabola, 100, 102, 104, 111 parabolic type, 158, 195 parametric, 180 plotting, 46 polar, 175 transcendental, 154, 163 trigonometric, 167 Exponential, equations, 163 type, 197 Extent, 51 Focus of, conic, 99 ellipse, 117, 120 hyperbola, 134, 137 parabola, 100 Formulas, algebraic, 2 differentiation, 230, 252, 255 integration, 248, 253, 258 logarithmic, 2 summary of, 303 trigonometric, 3 Frequency, 172 Functions, 216 algebraic, 225 amplitude of, 169 continuous, 235 decreasing, 236 explicit, 218 implicit, 218 increasing, 236 maximum value, 235 minimum value, 236 Digitized by Google 344 INDEX Functions, of variables, 44 period of, 169 periodic, 168 quadratic, 107 single-valued, 235 trigonometric, 7 Graph of equation, 46 Harmonic, conjugates, 206 motion, 170 ratio, 206 Higher plane curves, 154 Hooke's law, 80, 193 Hyperbola, applications, 150 asymptotes, 139 center, 137 conjugate, 141 conjugate axis, 137 conjugate diameters, 213 definition, 100, 134 difference between focal dis- tances constant, 148 directrix, 134, 137 eccentricity, 100, 135 equation, 135, 138, 143, 147 equation of tangent to, 234 equilateral, 142, 150 focus, 134, 137 general equation, 146, 155 latus rectum, 137 principal axis, 137 rectangular, 142 transverse axis, 137 vertices, 137 Hyperbolic, paraboloid, 289 type, 159, 197 Hyperboloids, 285 Hypocycloid, 183 Imaginary, circle, 87 ellipse, 123 number, 47 Inclination of line, 25 Increments, 216 Infinite, variable becomes, 221 Initial line, 30 Integral, indefinite, 247 Integration, 246 constant of, 247 formulas, 248, 253, 258 methods of, 248 Intercepts, 49, 283 Involute of circle, 186 Latus rectum of, ellipse, 120 hyperbola, 137 parabola, 102 Least squares, 193 Lemniscate, 163 Iimacpns of Pascal, 179 Limits, 220 theorems of, 221 Linear equations, 65, 191 Lines, applications, 79 directed, 8 direction cosines, 267 general equation, 65 inclination, 25 initial, 30 in polar coordinates, 78 in space, 298 intercept equation, 63 normal equation, 64 parallel, 27 perpendicular, 27 point direction equation, 299 point slope equation, 59 Digitized by Google INDEX 345 Lines, polar equation, 78 projection equation, 298 slope, 25, 65 slope intercept equation, 61 systems, 71 two point equation, 62, 300 line segment, 9 addition and subtraction of, 10, 11 division of, 19, 23, 264 magnitude of, 9 numerical value of, 9 projection of, 266 value of, 9, 16 Iituus, 180 Loci, algebraic, 154 composite, 53, 77 in space, 274 of equations, 45, 55 of points, 45 of polar equations, 175 through intersection of loci, 75 transcendental, 154 Logarithmic, equations, 165 paper, 200 Logarithms, formulas of, 2 M Major axis of ellipse, 120 Maximum, 235, 237 test of, 238 Method of least squares, 193 Minimum, 235, 237 test of, 238 Minor axis of ellipse, 120 N Newton's law of cooling, 167 Normals, 223, 292 Null circle, 87 Number, imaginary, 47 O Oblate spheroid, 285 Oblique coordinates, 15 Ordinate, 13, 14 Origin of codrdinates, 9, 29 Orthogonal projection, 266 Parabola, applications, 113 axis, 102 construction, 112 cubical, 158 definition, 100 diameters, 213 directrix, 100 eccentricity, 100 equation, 100, 102, 104, 111 equation of tangent to, 234 focus, 100 general equation, 109, 155 latus rectum, 102 semi cubical, 158 vertex, 102 Parabolic type, 158, 195 Paraboloids, 288 Parameter, 180 Period of function, 169 Periodic, curve, 168 function, 168 Periodicity factor, 169 Phase angle, 174 Planes, angle between, 295 determinant equation , 292 equations, 292 general equation, 292 in space, 292 intercept equation, 294 Digitized by Google 346 INDEX Planes, normal equation, 293 normal to, 292 parallel to axes, 274 Point, circle, 87 ellipse, 123 imaginary, 47 initial, 8 locus of, 45 of inflection, 239 polar coordinates of, 12, 29, 269 rectangular coordinates of, 12, 261 spherical coordinates of, 270 terminal, 8 Polar, coordinates, 12, 29, 174* 269 equation of circle, 96 equation of ellipse, 129 equation of hyperbola, 147 equation of line, 78 equation of parabola, 111 Pole, definition, 29 Poles and polars, 206 properties of, 209 Powers of e, 5 Probability curve, 199 Projectile, path of, 114 Projection, orthogonal, 266 of curves, 280 Prolate spheroid, 285 Proper sine curve, 168 Q Quadratic function, 107 Quadrie surfaces, 284 R Radical axis, 94 Radius vector, 30 Rectangular system, 13 Reflector, 115 Revolution, surfaces of, 276 Rotation of axes, 34, 109, 127, 156 formula "for, 155 Ruled surfaces, 286, 290, 291 S Simple harmonic motion, 170 Sine curve, 168 Slope, definition, 25 formula for, 25 Spheres, 276 Spherical coordinates, 270 Spheroids, 285 Spirals, 178 Archimedes, 179 center of, 178 hyperbolic, 179 logarithmic, 178 parabolic, 179 Supplemental chords, 215 Surfaces, 274 cylindrical, 274 equations of, 274 of revolution, 276 quadrie, 284 ruled, 286, 290, 291 sections of, 279 trace, 279 Symmetry, 49, 175, 177, 282 algebraic properties, 50 axis of, 50 center of. 49 Tables, e* and e"*, 5 of logarithms, 308 of trigonometric functions,310 Tangents, 223 equations of, to conies, 234 Torus, 277 Digitized by Google INDEX -*\_ 347 Trace of surfaces, 279, 283 V Transcendental equations, 154, 163 Transformation of coordinates, 33 Value of ]ine ^^^ 9> 16 Translation of axes, 33, 107, 123, Variables, 44, 216 143,157 become infinite, 221 Transverse axis of hyperbola, 137 dependant, 46 Triangle, area of, 36, 39 functions of, 44 center of gravity, 40 independant, 46 Trigonometric equations, 167 Vectorial angles, 30 Trigonometry, formulas of, 3 Vertice8 o{> ^ m functions of, 7 hyperbola, 137 P 0001 ^ 183 parabola, 102 Trumpet, 180 U w Uniform circular motion, 170 Witch of Agnesi, 162 Digitized by Google zediby U( Digitized by Google Digitized by Google Digitized by Google Digitized by Google [ >, Digitized by Google Digitized by Google w TB 17163 111: s rS!B: - if; i!j # •■ alii : ir ; : ]!l;: •fiV ni •■ :ijl|s tHE UNIVERSITY OP CALIFORNIA LIBRARY V Digitized by Google